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数字和代数

基本代数 - 表达式的展开和因式分解

Q.01

'展开式的通用项是\n\\[\\frac{6!}{p!q!r!} \\cdot a^{p} \\cdot(2 b)^{q} \\cdot(3 c)^{r}=\\frac{6!}{p!q!r!} \\cdot 2^{q} \\cdot 3^{r} \\cdot a^{p} b^{q} c^{r}\\]\n其中 \ \\quad p+q+r=6, p \\geqq 0, q \\geqq 0, r \\geqq 0 \\n(a) \ a^{3} b^{2} c \ 项的系数是, 当 \ p=3, q=2, r=1 \ 时为\n\\\frac{6!}{3!2!1!} \\cdot 2^{2} \\cdot 3^{1}=720\\n(b) \ a^{4} c^{2} \ 项的系数是, 当 \ p=4, q=0, r=2 \ 时为\n\\\frac{6!}{4!0!2!} \\cdot 2^{0} \\cdot 3^{2}=135\'

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Q.02

'求解以下展开式中指定项的系数。(1) (2x-y-3z)^6 [xy^3 z^2] (2) (1+x+x^2)^10 [x^4] (3) (x+1/x^2+1)^5 [常数项]'

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Q.03

'请找出以下表达式中特定项的通项公式和系数。'

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Q.04

'(1) \\((x+2-i)(x+2+i)\\)(2) \\((3 x-17)(2 x-9)\\)'

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Q.05

'请解决以下方程:'

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Q.06

'展开表达式 \\( (a+b+c)^{n} \\) 的一般项是\n\\\frac{n!}{p!q!r!} \\alpha^{p} b^{q} c^{r}\\n其中 \ p+q+r=n \'

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Q.07

'(2) (解 1) α^{3}+β^{3}+γ^{3}=(α+β+γ){α^{2}+β^{2}+γ^{2}-(αβ+βγ+γα)}+3αβγ =2 \\cdot(4-0)+3\\cdot4=20'

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Q.08

'(2) 展开式的通项是 frac10!p!q!r!cdot1pcdotxqcdotleft(x2right)r=frac10!p!q!r!cdotxq+2r\\frac{10!}{p!q!r!} \\cdot 1^{p} \\cdot x^{q} \\cdot\\left(x^{2}\\right)^{r}=\\frac{10!}{p!q!r!} \\cdot x^{q+2 r}\n其中 p+q+r=10quadcdotscdots(1),pgeqq0,qgeqq0,rgeqq0p+q+r=10 \\quad\\cdots\\cdots (1), p \\geqq 0, q \\geqq 0, r \\geqq 0。\nx4x^{4} 的项是在 q+2r=4q+2 r=4 时,即 q=42rq=4-2r。'

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Q.09

'求出数列 \ \\left\\{a_{n}\\right\\} \ 的通项公式。'

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Q.10

'展开以下多项式。'

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Q.11

'和符号 \ \\Sigma \, \ \\Sigma \ 的性质\n和符号 \ \\Sigma \\n\\n\\sum_{k=1}^{n} a_{k}=a_{1}+a_{2}+a_{3}+\\cdots \\cdots+a_{n}\n\\n这个性质中的 \ p, q \ 是与 \ k \ 无关的常数。\n\\[\n\\sum_{k=1}^{n}\\left(p a_{k}+q b_{k}\\right)=p \\sum_{k=1}^{n} a_{k}+q \\sum_{k=1}^{n} b_{k}\n\\]\n数列和公式中的 \ c, r \ 是与 \ n \ 无关的常数。\n\\[\n\egin{aligned}\n\\sum_{k=1}^{n} c & =n c \\\\ \n特别是 \\\\ \n\\sum_{k=1}^{n} 1=n \\\\ \n\\sum_{k=1}^{n} k & =\\frac{1}{2} n(n+1) \\\\ \n\\sum_{k=1}^{n} k^{2} & =\\frac{1}{6} n(n+1)(2 n+1) \\\\ \n\\sum_{k=1}^{n} k^{3} & =\\left\\{\\frac{1}{2} n(n+1)\\right\\}^{2} \\\\ \n\\sum_{k=1}^{n} r^{k-1} & =\\frac{1-r^{n}}{1-r} \\\\( r \\neq 1) \n\\end{aligned}\\]\n'

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Q.12

'二项式定理的应用问题'

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Q.13

'(A + B)(A - B)'

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Q.14

'(1) 6x² + 3 (2) −x⁸ + 3x³ − 1 (3) 2(x² + 1)'

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Q.15

'将给定文本翻译成多种语言。'

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Q.16

'(A + B)的平方加(A - B)的平方'

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Q.17

'请简化以下连分数:\n\\n\\frac{1}{1+\\frac{1}{1+\\frac{1}{x+1}}}\n\'

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Q.18

'(3) (2)求得的圆方程展开整理后为:x^2 - mx + y^2 - (m^2 + 2)y = 0。将y = x^2代入得x^2 - mx + x^4 - (m^2 + 2)x^2 = 0,即x(x + m)(x^2 - mx - 1) = 0。因此x = 0, -m, α, β。因此,抛物线y = x^2和(2)求得的圆A、B、O除外没有其他共有点的充要条件是x = -m是方程x(x^2 - mx - 1) = 0的根。'

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Q.19

'请在复数范围内因式分解以下二次方程:\n1. x^{2}+4 x+5\n2. 6 x^{2}-61 x+153'

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Q.20

'(2) 1 + 3x + 5x^{2} + ... + (2n - 1)x^{n - 1}'

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Q.21

'将给定文本翻译成多种语言。'

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Q.22

'(1) (x2)(x2+x+2)(x-2)(x^{2}+x+2)\n(2) (x+1)(x+2)(x3)(x+1)(x+2)(x-3)\n(3) (x1)2(x2+2x+3)(x-1)^{2}(x^{2}+2x+3)\n(4) (x+1)(x3)(x2+2)(x+1)(x-3)(x^{2}+2)\n(5) (3x+1)(4x23x+1)(3x+1)(4x^{2}-3x+1)\n(6) (x1)(2x+1)(x2+x1)(x-1)(2x+1)(x^{2}+x-1)'

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Q.23

'求P(x)=x³-4x²+x-7在x=-2時的餘數'

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Q.24

'(1) 由于解为 \ \\alpha, \eta \,得到'

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Q.25

'将P(x)除以(x+1)^{2}(x-2),商为Q(x),余数为R(x),则成立以下等式。'

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Q.26

'(2)\\n\\\\(\\n\\\egin{aligned}\\nS & =1+3 x+5 x^{2}+\\\\cdots \\\\cdots+(2 n-1) x^{n-1} \\\\\\nx S & =\\\\quad x+3 x^{2}+\\\\cdots \\\\cdots+(2 n-3) x^{n-1}+(2 n-1) x^{n}\\n\\end{aligned}\\n\\\\)\\n\\n边边引くと\\\\( (1-x) S=1+2\\left(x+x^{2}+\\\\cdots \\\\cdots+x^{n-1}\\right)-(2 n-1) x^{n} \\)\\n\\nよって, \\\\x \\neq 1 \\\\ のとき\\\\(\\n\\\egin{aligned}\\n(1-x) S & =1+2 \\\\cdot \\\\frac{x\\left(1-x^{n-1}\\right)}{1-x}-(2 n-1) x^{n} \\\\\n& =\\\\frac{1-x+2\\left(x-x^{n}\\right)-(2 n-1) x^{n}(1-x)}{1-x} \\\\\n& =\\\\frac{1+x-(2 n+1) x^{n}+(2 n-1) x^{n+1}}{1-x}\\n\\end{aligned}\\n\\\\)\\n\\nゆえに \\\\ S=\\\\frac{1+x-(2 n+1) x^{n}+(2 n-1) x^{n+1}}{(1-x)^{2}} \\\\n\\\\( x=1 \\text{ のとき } \\quad \egin{aligned}\\nS & =1+3+5+\\\\cdots \\\\cdots+(2 n-1)=\\\\sum_{k=1}^{n}(2 k-1) \\\\\n& =2 \\\\cdot \\\\frac{1}{2} n(n+1)-n=n^{2}\\n\\end{aligned}\\n\\)'

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Q.27

'确定常数a、b、c、d的值,使等式( x + a y - 3)(2 x - 3 y + b) = 2 x^{2} + c x y - 6 y^{2} - 4 x + d y - 6成为关于x、y的恒等式。'

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Q.28

'\\[ 3(a x+2 b y)-(a+2 b)(x+2 y) \\]\n\\[=3 a x+6 b y-(a x+2 a y+2 b x+4 b y) \\]\n\\[=2(a x-a y-b x+b y) \\]\n\\[=2\\{ a(x-y)-b(x-y) \\} \\]\n\\[=2(a-b)(x-y) \\]\n\ a>b, x>y 所以, a-b>0, x-y>0 \\n\\[2(a-b)(x-y)>0 \\]\n\因此 \\n\\[(a+2 b)(x+2 y)<3(a x+2 b y) \\]'

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Q.29

'另 x^{3/2} + x^{-3/2} = (x^{1/2} + x^{-1/2})^3 - 3x^{1/2}x^{-1/2}(x^{1/2} + x^{-1/2})'

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Q.30

'(A - B)(A^2 + AB + B^2)'

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Q.31

'求解给定展开式中指定项的系数。(1) (2 x+3 y)^{4} [x^{2} y^{2}] (2) (3 a-2 b)^{5} [a^{2} b^{3}]'

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Q.32

'求解展开式的通项表达式。'

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Q.33

'设f(y)=y2+ya9f(y)=y^{2}+y-a-9,则关于轴的情况为3<\x0crac12<3-3<-\x0crac{1}{2}<3。由f(3)>0f(3)>0可得a<3a<3,且f(3)>0f(-3)>0可得a<3a<-3。综上得到\x0crac374<a<3-\x0crac{37}{4}<a<-3。'

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Q.34

'练习问题:求展开式中的x₁^p, x₂^p, ..., xᵣ^p的系数。'

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Q.35

'使用二项式定理展开(a+b)ⁿ。'

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Q.36

'数学 I\n267\n\\[\egin{aligned} y_{1}+y_{2} & =\\triangle \\mathrm{OAP}-\\int_{0}^{1}(-3x^{2}+3)dx+2y_{1} \\& =\\frac{1}{2} \\cdot 1 \\cdot 3p+3 \\int_{0}^{1}(x^{2}-1)dx+2 \\cdot \\frac{1}{2}(2-p)^{3} \\& =\\frac{3}{2}p+3\\left[\\frac{x^{3}}{3}-x\\right]_{0}^{1}+(2-p)^{3} \\& =\\frac{3}{2}p-2+(2-p)^{3} \\& =-p^{3}+6p^{2}-\\frac{21}{2}p+6 \\end{aligned}\\]'

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Q.37

'(2)给定方程组的解为 \ \\alpha, \eta \ ,因此'

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Q.38

'练习79册302页y = a x ^(3)-2 x上的点(t,a t3-2 t)和原点的距离的平方为t ^(2)+(a t3- 2 t) ^(2)=a ^(2)t6-4 a t4+5 t2'

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Q.39

'请说明使得上述等式为0的条件。'

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Q.40

'对于实数t,考虑两点P(t, t^{2})和Q(t+1, (t+1)^{2})。'

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Q.41

'(2) 由 f(a)=f(a+1) 推导出 a^{3}-3 a=(a+1)^{3}-3(a+1)'

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Q.42

'求出展开式中指定项的系数。(1) (x^2+2y)^5 [x^4 y^3] (2) (x^2-2/x)^6 [x^6, 常数项]'

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Q.43

'已知 19x^{3} 的系数为1的3次方程式Q(x)除以x-1的余数为-1,除以x-2的余数为8。'

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Q.44

'关于序列 \ \\{a_{n}\\} \,其中总和从第1项到第n项为 \ S_{n}=2 n^{2}-n \,请回答以下问题:\n1. 求一般项 \ a_{n} \。\n2. 求和 \ a_{1}+a_{3}+a_{5}+ \\ldots \\ldots+a_{2 n-1} \。'

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Q.45

''

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Q.46

'解决以下方程式。'

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Q.47

'请检查以下等式是否恒等:\n(1) (x-1)^{2}=x^{2}+1\n(2) (a+b)^{2}+(a-b)^{2}=2(a^{2}+b^{2})\n(3) \\frac{2 x+1}{2 x-1} \\times \\frac{4 x^{2}-1}{(2 x+1)^{2}}=1\n(4) \\frac{1}{3}\\left(\\frac{1}{x+1}-\\frac{1}{x+3}\\right)=\\frac{1}{(x+1)(x+3)}'

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Q.48

''

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Q.49

'展开左侧的复杂表达式,以示变为简单的右侧表达式。\n(2),(3)由于左侧和右侧同样复杂,因此对左侧和右侧进行变形,以示它们成为相同的表达式。'

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Q.50

'将2x²-8分解为(x²+4)(x²-2)'

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Q.51

'演习20 III \ \\Rightarrow \ 本册 \ p. 471 \\n(1) 假设1或2是大骰子的结果为 \ X \\n\n\\[ x_{n}=1 \\cdot X+(-1) \\cdot(n-X)=2 X-n \\]\n\n由于 \ X \ 符合二项分布 \\( B\\left(n, \\frac{1}{3}\\right) \\), 所以, \ X \ 的均值 \\( E(X) \\) 和方差 \\( V(X) \\) 是 \\( E(X)=\\frac{n}{3} \\), \\( V(X)=n \\cdot \\frac{1}{3} \\cdot \\frac{2}{3}=\\frac{2}{9} n \\)。\n因此, \ x_{n} \ 的均值 \\( E\\left(x_{n}\\right) \\) 和方差 \\( V\\left(x_{n}\\right) \\) 为\n\\[\egin{aligned}\nE\\left(x_{n}\\right) & =E(2 X-n)=2 E(X)-n \\\\ & =2 \\cdot \\frac{n}{3}-n=-\\frac{n}{3} \\\\nV\\left(x_{n}\\right) & =V(2 X-n)=2^{2} V(X)=\\frac{8}{9} n\n\\end{aligned}\n\\]\n(2) 由于 \\( V\\left(x_{n}\\right)=E\\left(x_{n}^{2}\\right)-\\left\\{E\\left(x_{n}\\right)\\right\\}^{2} \\) 所以\n\\[E\\left(x_{n}^{2}\\right)=V\\left(x_{n}\\right)+\\left\\{E\\left(x_{n}\\right)\\right\\}^{2}=\\frac{8}{9} n+\\left(-\\frac{n}{3}\\right)^{2}=\\frac{1}{9} n(n+8)\\]\n\n(3) 由于 \\( S=\\pi\\left(x_{n}{ }^{2}+y_{n}{ }^{2}\\right) \\), 所以, \ S \ 的均值 \\( E(S) \\) 为\n\n\\[E(S)=\\pi\\left\\{E\\left(x_{n}{ }^{2}\\right)+E\\left(y_{n}{ }^{2}\\right)\\right\\}\\]\n\n现在, 让我们计算 \ y_{n} \ 的均值 \\( E\\left(y_{n}\\right) \\) 和方差 \\( V\\left(y_{n}\\right) \\)。假设1是小骰子的结果为 \ Y \, 那么\n\ y_{n}=2 Y-n \\n\ Y \ 也符合二项分布 \\( B\\left(n, \\frac{1}{6}\\right) \\), 类似于第(1)部分\n\\[ \egin{array}{l}\nE\\left(y_{n}\\right)=2 \\cdot \\frac{n}{6}-n=-\\frac{2}{3} n \\\\\nV\\left(y_{n}\\right)=2^{2} \\cdot n \\cdot \\frac{1}{6} \\cdot \\frac{5}{6}=\\frac{5}{9} n\n\\end{array} \\]\n类似地, 根据第(2)部分\n\\[\egin{aligned}\nE\\left(y_{n}^{2}\\right) & =\\frac{5}{9} n+\\left(-\\frac{2}{3} n\\right)^{2}=\\frac{1}{9} n(4 n+5) \\\\\n\\text { 因此 } \\quad E(S) & =\\pi\\left\\{\\frac{1}{9} n(n+8)+\\frac{1}{9} n(4 n+5)\\right\\} \\\\\n& =\\frac{1}{9} n(5 n+13) \\pi\n\\end{aligned}\\]'

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Q.52

'数学I即(α-1)(β-1)(γ-1)=0,所以α,β,γ至少有一个是1。'

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Q.53

'求多项式 x^2020+x^2021 除以多项式 x^2+x+1 的余数。'

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Q.54

'(2) 令t=x+1/x,则证明利用数学归纳法得到x^n+1/x^n将成为t的n次方程。'

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Q.55

'计算给定的指数'

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Q.56

'设k为实数。 对于三次方程f(x)=x^{3}-kx^{2}-1,设方程f(x)=0的三个解为α,β,γ。 g(x)是一个三次方程,其x^{3}的系数为1,并且方程g(x)=0的三个解为αβ,βγ,γα。\n(1) 请用α,β,γ表达g(x)。\n(2) 求解具有共同解的两个方程f(x)=0和g(x)=0的k值。'

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Q.57

'在练习本册中(第35页), 如果将P的三次项的系数记为a, b, c为常数, 则P = (x+1)^2(ax+b), P-4 = (x-1)^2(ax+c)。'

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Q.58

'将给定的文本翻译成多种语言。'

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Q.59

'300'

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Q.60

'练习 56 (1) (上半部分) P_1=α+β=(1+√2)+(1-√2)=2 又αβ=(1+√2)(1-√2)=-1 因此 P_2=α^2+β^2=(α+β)^2-2αβ=2^2-2(-1)=6 (下半部分) [1] 当n=1时,P_1=2,当n=2时,P_2=6 因此,当n=1,2时,P_n 是非4倍数的偶数。 [2] 假设n=k, k+1时,P_n 是非4倍数的偶数。'

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Q.61

'首项为 a,公差为 d,将首项到第 n 项求和为 S_{n},则 S_{5}=125,S_{10}=500,因此1/2・5{2a+(5-1)d}=125,1/2・10{2a+(10-1)d}=500,得出 a+2d=25 ... (1),2a+9d=100 ... (2)。解方程组 (1),(2) 可得 a=5,d=10'

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Q.62

'关于整式f(x)=x^{4}-x^{2}+1,回答以下问题。'

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Q.63

'(1 + xi)(3 - i)这个表达式是实数还是纯虚数取决于 x 的值。'

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Q.64

'(3u-v)(-u³+3u-v) < 0'

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Q.65

'(1) left(x2+2yright)5 \\left(x^{2}+2 y\\right)^{5} 的展开式的一般项是5mathrmCrleft(x2right)5r(2y)r=5mathrmCrcdot2rx102ryr _{5} \\mathrm{C}_{r}\\left(x^{2}\\right)^{5-r}(2 y)^{r}={ }_{5} \\mathrm{C}_{r} \\cdot 2^{r} x^{10-2 r} y^{r} x4y3 x^{4} y^{3} 的项是r=3r=3时,其系数是'

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Q.66

'展开以下表达式:(a+b)³和(a-b)³'

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Q.67

''

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Q.68

'确定常数 a、b 和 c 的值,使得该等式对于 x 是恒等式。'

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Q.69

'求解展開式(a+2b+3c)^{6}中a^{3} b^{2} c項的系数,以及a^{4} c^{2}項的系数。'

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Q.70

'数列 {P_{n}} 被定义如下。'

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Q.71

'请列举数字版图表式参考书的三个基本功能。'

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Q.72

'等差数列的证明'

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Q.73

'练习,分解下列方程式。'

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Q.74

'是否能将 P 分解为 x、y 的一次式的乘积取决于 α、β 不是 y 的一次式。'

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Q.75

'展开以下表达式。(1) (a+2 b)^{7} (2) (2 x-y)^{6} (3) (2 m+n/3)^{6}'

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Q.76

'利用二项式定理,证明下面的等式。'

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Q.77

'因为抛物线y=x^2+bx+c的顶点在直线y=x上,所以顶点的坐标可以设为(k, k)。因此,抛物线的方程为y=(x-k)^2+k即y=x^2-2kx+k^2+k。抛物线(1)和抛物线y=-x^2+4的交点的x坐标为x^2-2kx+k^2+k=-x^2+4即2x^2-2kx+k^2+k-4=0的实数解。(1),(2)有两个不同的交点,所以设(3)的判别式为D,则D>0。计算D/4=(-k)^{2}-2(k^{2}+k-4)=-k^{2}-2k+8,由此-k^{2}-2k+8>0,即k^{2}+2k-8<0解得-4<k<2.在这种情况下,将两个交点的x坐标记为α、β(α<β),α、β是(3)的解,因此有α+β=k,αβ=(k^{2}+k-4)/2。因此,(β-α)^{2}=(α+β)^{2}-4αβ=k^{2}-2(k^{2}+k-4)=-k^{2}-2k+8=-(k+1)^{2}+9。'

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Q.78

'考虑当数学B329 n=k+2时'

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Q.79

'请证明函数的恒等式。'

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Q.80

'转化递推式, 数学归纳法转化递推式\n- 相邻两项 \\( a_{n+1} = p a_{n} + q \\(p \\neq 1) \\) 对于满足 \ \\alpha = p \\alpha + q \ 的 \ \\alpha \\n\\[\na_{n+1} - \\alpha = p\\left(a_{n} - \\alpha\\right) \n\\]\n- 相邻三项 \ p a_{n+2} + q a_{n+1} + r a_{n} = 0 \ \ p x^{2} + q x + r = 0 \ 的解为 \ \\alpha, \eta \ 那么\n\\[\na_{n+2} - \\alpha a_{n+1} = \eta\\left(a_{n+1} - \\alpha a_{n}\\right)\n\\]\n数学归纳法\n要证明命题 \ P \ 关于自然数 \ n \ 对于所有自然数成立的步骤是\n[1] 证明当 \ n=1 \ 时 \ P \ 成立。\n[2] 假设当 \ n=k \ 时 \ P \ 成立, 证明当 \ n=k+1 \ 时 \ P \ 也成立。'

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Q.81

'(3) 认为存在实数p, q, r, s, t, u满足方程x^{2}+y^{2}-5=(p x+q y+r)(s x+t y+u)。展开右边后,x^2的系数为p s,比较两边的x^2系数得到ps=1。因此,必须满足p不等于0,s不等于0。'

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Q.82

'设a为实常数,考虑两个圆C1:x^{2}+y^{2}=4,C2:x^{2}-6x+y^{2}-2ay+4a+4=0'

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Q.83

'请将以下表达式进行因式分解。'

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Q.84

'求多项式P(x)=4x32x25x+3P(x)=4 x^{3}-2 x^{2}-5 x+3被一次式除时的余数。'

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Q.85

'使用因式定理对以下方程进行因式分解。'

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Q.86

'请计算以下表达式的系数。(6) x^6-12x^5+60x^4-160x^3+240x^2-192x+64'

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Q.87

'拓展 51:因式分解2次2項式(使用解公式)'

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Q.88

'组立除法\n将三次方程式 P(x)=ax3+bx2+cx+d P(x)=a x^{3}+b x^{2}+c x+d 除以一次方程式 xk x-k 得到商式 Q(x)=lx2+mx+n Q(x)=l x^{2}+m x+n 和余数 R R 。\n这个商的系数 l,m,n l, m, n 和余数 R R 可以通过下面的方法求得,这个方法称为组立除法。\n\n正明 根据除法等式 P(x)=(xk)Q(x)+R P(x)=(x-k) Q(x)+R 可得\n\\[\na x^{3}+b x^{2}+c x+d=(x-k)\\left(l x^{2}+m x+n\\right)+R\n\\]\n这个等式是关于 x x 的恒等式。\n将右边展开整理得到\n\\[\na x^{3}+b x^{2}+c x+d=l x^{3}+(m-l k) x^{2}+(n-m k) x+(R-n k)\n\\]\n比较两边系数得到\n\\na=l, \\quad b=m-l k, c=n-m k, d=R-n k\n\\]\n因此\n\\[\nl=a, \\quad m=b+l k, \\quad n=c+m k, \\quad R=d+n k\n\'

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Q.89

'求展开式中带有项的系数'

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Q.90

'请检查以下等式是否恒等式。'

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Q.91

'设k为常数。 当(a+kb+c)^{5}的展开式中 a^{2}bc^{2}的系数为60时,求k的值。另外,求ac^{4}的系数。'

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Q.92

''

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Q.93

'请对给定方程进行因式分解。'

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Q.94

'恒等式的系数确定(1)...系数比较法'

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Q.95

'[7!/(3!2!2!)]'

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Q.96

'使用二项式定理展开(a+b)^{4},并求得各项的系数。'

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Q.97

'数列 {a_{n}}: 1, 3, 8, 19, 42, 89, 的差分序列为 {b_{n}}。若数列 {b_{n}} 的差分序列是等比数列,则\n(1) 求数列 {b_{n}} 的一般项。\n(2) 求数列 {a_{n}} 的一般项。《基本例题 19'

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Q.98

'使用二项式定理,求下列表达式的展开式。'

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Q.99

'请求解以下表达式展开式中括号内项的系数。'

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Q.00

'为了使等式x31=a(x1)(x2)(x3)+b(x1)(x2)+c(x1)x^{3}-1=a(x-1)(x-2)(x-3)+b(x-1)(x-2)+c(x-1)对于所有的xx成立, 确定常数a,b,ca, b, c的值。'

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Q.01

'确定常数a和b的值,使得下列多项式能被给定的表达式整除:'

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Q.02

'\ 30 \\quad \\frac{7}{3} x^{2}-x-\\frac{1}{3} \'

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Q.03

'请因式分解以下表达式。'

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Q.04

'部分分数分解'

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Q.05

'基础 61:高次方程式的解法(1)- 利用因式分解'

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Q.06

'求展开式中 [ ] 内项的系数。'

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Q.07

'在高次多项式的因式分解中,找到使 P(k)=0 成立的 k,然后使用因式定理。 这里我们将讨论如何找到使 P(k)=0 成立的 k。'

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Q.08

'展开式中 (a+b+c)^{n} 的系数是什么?'

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Q.09

'求展開式中的 [ ] 項的係數。'

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Q.10

'找出满足以下条件的多项式 A 和 B。'

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Q.11

'基础 45: 在复数范围内对二次方程进行因式分解'

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Q.12

'(a+b+c)^{5} [a b^{2} c^{2}]的展開式中,[ ]內項的係數是多少?'

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Q.13

'二项式定理'

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Q.14

'请检查以下方程是否恒等式。'

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Q.15

'在数学 I 中,我们学习了因式分解,并学习了如何利用它来解二次方程。在这里,我们将利用因式定理来考虑如何解 n 次以上的方程。'

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Q.16

'设二次方程 (x1)(x2)+(x2)(x3)+(x3)(x1)=0(x-1)(x-2)+(x-2)(x-3)+(x-3)(x-1)=0 的两个解为 alpha,\eta\\alpha, \eta,求下列式子的值。 (1) alpha\eta\\alpha\eta (2) (1alpha)(1\eta)(1-\\alpha)(1-\eta) (3) (alpha2)(\eta2)(\\alpha-2)(\eta-2)'

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Q.17

'将x^2+1/(x^2-1)转换为4(x^2-1)+1/(x^2-1)+4进行思考。'

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Q.18

'请使用二项式定理展开(a+b)^4。'

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Q.19

'请确定常数a、b、c的值,使下面的等式成立,其中x为恒等式(1) \\frac{4 x+5}{(x+2)(x-1)}=\\frac{a}{x+2}+\\frac{b}{x-1}(2) \\frac{3 x+2}{x^{2}(x+1)}=\\frac{a}{x}+\\frac{b}{x^{2}}+\\frac{c}{x+1}'

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Q.20

'已知 B = x^2 + x - 3, Q = 4x - 1, R = 13x - 5,求 A。'

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Q.21

'(2) \\( (x-1)(x+1)(x-2)(x-3) \\)'

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Q.22

'展开以下表达式。'

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Q.23

'曲线 C 和直线 l 的共享点的 x 坐标是,方程 x^{3}+2 x^{2}-4 x-8=0 的实数解。左边有 x+2 作为因数,因此因式分解得到 (x+2)^{2}(x-2)=0,即 x=2,-2,因此,曲线 C 和直线 l 的共享点中,除了接点之外的点的 x 坐标是2。'

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Q.24

'考虑数列{a_n}从第一个数到第五个数,对于n=1,2,3,4,有a_{n+1}=a_{n}+A×10^{n}....对于所有自然数n成立(1),这时,a_{n+2}=a_{n}+B×10^{n}....(2) 成立。a_{1}=11, a_{2}=101, (2)表示,当n为E时,a_{n}是11的倍数,a_{n}是11的倍数时,n是F。'

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Q.25

''

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Q.26

'展开下列表达式。'

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Q.27

'请在复数范围内因式分解以下二次方程:\n(1) \x^{2}-3 x-3 \\n(2) \ 2 x^{2}+4 x-1 \\n(3) \ 2 x^{2}-3 x+2 \'

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Q.28

'设数列{a_{n}},定义b_{n}=\\frac{a_{1}+a_{2}+\\cdots \\cdots+a_{n}}{n}'

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Q.29

'(3) \\( (2 x+1)\\left(3 x^{2}-x+2\\right) \\)'

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Q.30

'求展开式中的 [ ] 中项的系数。6 (1) (x+y+z)^{8}[x^{2} y^{3} z^{3}] (2) (x-y-2 z)^{7} [x^{3} y^{2} z^{2}]'

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Q.31

'证明当a+b+c=0时,a^{2}-b c=b^{2}-c a成立。'

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Q.32

'(1) (x-1)(x+1)(x+3)'

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Q.33

'使用积和公式'

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Q.34

'在数学I中,我们学习了二次式。在数学II中,我们将学习处理三次式等高次数的表达式。因此,让我们首先学习三次式的展开和因式分解。'

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Q.35

'求展开式中的 x^4 项的系数。'

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Q.36

''

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Q.37

'将三次多项式展开并因式分解'

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Q.38

'设2次方程2x²-3x+5=0的两个解为α、β,则以α²、β²为解的二次方程是?'

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Q.39

'使用解的公式进行512元的二次方程的因式分解。'

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Q.40

'确定常数a,b的值,使得下面的等式对于x成立:'

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Q.41

'使用二项式定理,找出以下表达式的展开式。'

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Q.42

'请因式分解以下方程:\\(x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)\\)。'

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Q.43

'[x^3]的系数是多少?'

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Q.44

'如何从递推式中求得通项公式。\\n解下列递推式,求得数列通项公式:\\n\\n1. 等差数列类型\\n\ a_{n+1}=a_{n}+d \\\n\ [d \ 为常数 \\])\\n\\n2. 等比数列类型\\n\ a_{n+1}=r a_{n} \\\n\ [r \ 为常数 \\])\\n\\n3. 差分数列类型\\n\\( a_{n+1}=a_{n}+f(n) \\)\\n\\( [ f(n) 为差分数列通项公式 \\])\\n\\n此外,\\n\ a_{n+1}=p a_{n}+q\\\n\ p \ 和 \ q \ 为常数, \\( p \\neq 1, q \\neq 0 \\)\\n的递推式形式,并解出数列的通项公式。'

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Q.45

'基础 59:高次多项式的因式分解'

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Q.46

''

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Q.47

'求15^4(1+x+x^2)^{8}展开式中x^{11}的系数。'

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Q.48

'请因式分解以下方程式。'

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Q.49

'请计算在以下情况下,A除以B的商和余数。'

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Q.50

'分解下列方程式。'

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Q.51

'数列 {a_{n}}: 1,3,8,19,42,89, \\cdots \\cdots 的差分序列为 {b_{n}}。 当数列 {b_{n}} 的差分序列为等比数列时:(1)求数列 {b_{n}} 的一般项。(2)求数列 {a_{n}} 的一般项。'

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Q.52

'求解展開式'

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Q.53

'展开和因式分解的表达式'

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Q.54

'当 a=2 时,(x-2y+1)(x+y+1),当 a=-5/2 时,(x-2y-2)(x+y-1/2)'

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Q.55

'训练13 求下列等比数列的和。 (1) 第一项为4,公比为1/2,项数为7 (2) 数列3,-3,3,-3,...,项数为n (3) 数列18,-6,2,...,项数为n'

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Q.56

'求调和数列{an}的通项,其中第2项为1,第5项为1/13。'

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Q.57

'解4次方程式x^4+8x^3+20x^2+16x-12=0。'

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Q.58

'除了搅拌之外,给出溶解固体以增快溶解速度的两种方法。假设水和固体的量保持不变。'

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Q.59

'图1显示了一个教室的座位表。共有9个座位,学生全都面向黑板坐着。为了避免前后左右的座位连在一起,座位是确定的。例如,当为座位编号,如果有学生坐在①号座位,其他学生就不能坐在2号和4号座位。请回答以下问题。(1) A、B、C、D、E五位学生坐下时,有多少种确定座位的方式?(2)A、B、C、D四位学生坐下时,有多少种确定座位的方式?(3)A、B、C三位学生坐下时,有多少种确定座位的方式?'

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Q.60

'求1+x+x^2+⋯+x^n的和。'

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Q.61

'將給定的文本翻譯成多種語言。'

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Q.62

'从想成为的自己开始逆向思维。'

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Q.63

'针对数列{an},请回答以下问题:(1)求数列{an^2 + bn^2}的通项公式。并求lim_{n->∞} (an^2 + bn^2)。 (2)证明lim_{n->∞} an = lim_{n->∞} bn = 0。并求∑_{n=1}^{∞} an,∑_{n=1}^{∞} bn。'

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Q.64

'近似值与近似表达式'

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Q.65

''

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Q.66

'从单词mathematics中随机选取4个字母,形成排列的数量。'

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Q.67

'请因式分解以下表达式。'

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Q.68

'在列出PR NAGOYAJO的8个字符的所有排列中,同时包含AA和OO的排列有多少个,不相邻字符的排列有多少个。'

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Q.69

'请因式分解以下方程:'

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Q.70

'请计算以下表达式。'

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Q.71

'将4个A,5个B和2个C分成组合的方法是C_9^5×C_4^2。'

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Q.72

'2 x^{2}-6 x+4'

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Q.73

'将3名学生放入A的方法有 C_9^3 种'

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Q.74

'请将3x ^ {2} -5x +1补全为完全平方。'

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Q.75

'19(1)\\((x+y-1)\\left(x^{2}-x y+y^{2}+x+y+1\\right)\\ (2)\\((x-2 y-z)\\left(x^{2}+4 y^{2}+z^{2}+2 x y-2 y z+z x\\right)'

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Q.76

'整理给定多项式的同类项。'

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Q.77

'展开下列表达式。'

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Q.78

'展开以下表达式。'

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Q.79

'(1) \\( 3(a+b)(b+c)(c+a) \\)\\n(2) \\( (a b+a+b-1)(a b-a-b-1) \\)'

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Q.80

'将以下表达式进行因式分解:2x2+5xy+2y23y22x^{2}+5xy+2y^{2}-3y-2'

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Q.81

'展开以下表达式。'

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Q.82

'请因式分解以下表达式。'

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Q.83

'(x-1)(x+2)(x-3)(x+4)+24的因式分解是什么?'

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Q.84

'因式分解以下表达式。'

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Q.85

'因此,所需排列的数量为\n\\[\n\egin{aligned}\n10080- & 24 \\times(30+30+30+20) \\\\\n& =10080-24 \\times 110=10080-2640 \\\\\n& =7440 \\text { (种) }\n\\end{aligned}\n\\]'

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Q.86

'10\n(1)\\((x-3)(3 x-1)\\)\n(2)\\((x+1)(3 x+2)\\)\n(3)\\((a+2)(3 a-1)\\)\n(4)\\((a-3)(4 a+5)\\)\n(5)\\((2 p+3 q)(3 p-q)\\)\n(6)\\((a x-b)(b x+a)\\)'

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Q.87

'请因式分解以下表达式。\n(1) x^{3}+3xy+y^{3}-1'

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Q.88

'整理多项式'

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Q.89

'计算以下表达式。'

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Q.90

'回答以下实数的子集问题。'

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Q.91

'请因式分解以下表达式:2x23xy+y2+7x5y+62x^{2}-3xy+y^{2}+7x-5y+6'

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Q.92

'将以下表达式进行因式分解。(1) (x+y)^{2}-4(x+y)+3 (2) 9 a^{2}-b^{2}-4 b c-4 c^{2} (3) (x+y+z)(x+3 y+z)-8 y^{2} (4) (x-y)^{3}+(y-z)^{3}'

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Q.93

'请根据 x 进行降幂排序。'

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Q.94

'展开以下表达式。'

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Q.95

'请因式分解以下方程:\n(1) 2 x^{3}+16 y^{3}\n(2) (x+1)^{3}-27'

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Q.96

'(4)\\((3 a-b)(9 a^{2}+3 a b+b^{2})\\)的展开是:'

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Q.97

'将给定的表达式进行因式分解。'

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Q.98

'展开表达式(2x + 3y + z)(x + 2y + 3z)(3x + y + 2z)后,求解xyz的系数。'

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Q.99

'给定字符串的排列总数是多少?'

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Q.00

'76 \\quad y=\\frac{1}{3}(x+1)(x-5)\n\\( \\left(y=\\frac{1}{3} x^{2}-\\frac{4}{3} x-\\frac{5}{3}\\right) \\)'

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Q.01

'10 \u3000 809 11 (1) \\\\ ( 2(x+2 y)(x^{2}-2 x y+4 y^{2}) \\) (2) \\\\ (x-2)(x^{2}+5 x+13) \\)'

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Q.02

'请将{1}/{3}x^{2}+2x+1完成平方。'

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Q.03

'展开公式'

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Q.04

'展开表达式(a+b+c+d)(p+q+r)(x+y),会得到多少项?'

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Q.05

'请计算以下表达式。'

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Q.06

'多项式的乘积展开,可以通过反复利用分配律来展开复杂的表达式。但是,如果继续进行计算而没有考虑因式分解的步骤,很容易陷入困境。在这里,我们按照优先级高低的顺序总结了找到因式分解步骤的方法。在考虑因式分解时,请注意这些要点。'

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Q.07

'设A=5x³ -2x² +3x +4,B=3x³ -5x² +3,计算以下结果:(1) A+B (2) A-B'

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Q.08

'请将-2 x^{2}+10 x-7完成平方。'

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Q.09

'请分解以下表达式。'

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Q.10

'请分解以下方程:x2+3xy+2y2+2x+3y+1x^{2}+3xy+2y^{2}+2x+3y+1'

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Q.11

'(3) \\((3 x+x^{3}-1)\\left(2 x^{2}-x-6\\right)\\)'

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Q.12

'请简化以下数学表达式。'

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Q.13

'3个集合的交集,并集\n交集A∩B∩C是属于A、B、C的元素集合的所有元素。\n并集A∪B∪C是属于至少一个A、B、C的元素集合的所有元素。\n关于3个集合的性质\n(1)\n\\[\n\egin{aligned}\nn(A∪B∪C)= & n(A)+n(B)+n(C) \\\\\n& -n(A∩B)-n(B∩C)-n(C∩A)+n(A∩B∩C)\n\\end{aligned}\n\\]\n(数量定理的扩展)\n(2) \\\overline{A∪B∪C}=\\overline{A} \\cap \\overline{B} \\cap \\overline{C}, \\overline{A∩B∩C}=\\overline{A} \\cup \\overline{B} \\cup \\overline{C} \\n(德摩根律的扩展)'

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Q.14

''

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Q.15

'展开表达式(2x+3y+z)(x+2y+3z)(3x+y+2z)并求出xyz的系数。'

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Q.16

'(例) 对于方程 x^2 - 2 xy + 2 y^2 = 13(x > 0,y > 0)'

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Q.17

'请分解以下表达式:\n\nx^3 - 8y^3 - z^3 - 6xyz'

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Q.18

''

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Q.19

'展开以下表达式。'

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Q.20

'请简化这个表达式。'

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Q.21

'(5) 展开以下表达式。(x+y+z)(x-y-z)'

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Q.22

'展开以下表达式。'

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Q.23

'方幂定理'

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Q.24

'对称表达式'

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Q.25

'展开以下表达式。'

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Q.26

'6(1)\\n\\[\\n\\\egin{aligned}\\n x^{2}-14 x+49 & =x^{2}-2 \\\\cdot x \\\\cdot 7+7^{2} \\\\\n & =(x-7)^{2} \\end{aligned} \\n\\]\\n(2)\\(a^{2}+12 a b+36 b^{2}=a^{2}+2 \\\\\\\\cdot a \\\\\\\\cdot 6 b+(6 b)^{2}\\)\\n\\[\\n=(a+6 b)^{2}\\n\\]\\n(3)\\(25 a^{2}-81=(5 a)^{2}-9^{2}\\)\\n\\[\\n=(5 a+9)(5 a-9)\\n\\]\\n(4)\\n\\[\\n\\\egin{aligned}\\n9 x^{2}-64 y^{2} & =(3 x)^{2}-(8 y)^{2} \\\\\\\\n & =(3 \oldsymbol{x}+8 \oldsymbol{y})(3 \oldsymbol{x}-8 \oldsymbol{y})\\n\\end{aligned}\\n\\]\\n(5)\\n\\[\\n\\text{5) } \egin{aligned}\\n& 25 x^{2}+40 x y+16 y^{2} \\\\\\\\n= & (5 x)^{2}+2 \\\\\\\\cdot 5 x \\\\\\\\cdot 4 y+(4 y)^{2} \\\\\\\\n= & (5 x+4 y)^{2}\\n\\end{aligned}\\n\\]\\n(6)\\n\\[\\n\\text{(6) } \egin{aligned}\\n& 9 a^{2}-42 a b+49 b^{2} \\\\\\\\n= & (3 a)^{2}-2 \\\\\\\\cdot 3 a \\\\\\\\cdot 7 b+(7 b)^{2} \\\\\\\\n= & (3 a-7 b)^{2}\\n\\end{aligned}\\n\\]\\n(7)\\n\\[\\n\\\egin{aligned}\\n x^{2}+5 x+6 & =x^{2}+(2+3) x+2 \\\\\\\\n x^{2}-7 x+12 & =x^{2}+(-3-4) x+(-3) \\\\\\\\n & =(x-3)(x-4)\\n\\end{aligned}\\n\\]\\n(8)\\n\\[\\n\\\egin{aligned}\\n x^{2}-7 x+12 & =x^{2}+(-3-4) x+(-3) \\\\\\\\n & =(x-3)(x-4)\\n\\end{aligned}\\n\\]'

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Q.27

'将给定方程转化为形式 y=a(x-p)^{2}+q(完成平方)。'

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Q.28

'请因式分解以下表达式。'

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Q.29

'请将以下方程分解因式。'

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Q.30

'12(1) \\((x-y)(2x+y-1)\\) (2) \\((x+y-3)(3x+y+2)\\) (3) \\((x+2y-1)(3x-y+2)\\) (4) \\((x+y-z)(x-2y+z)\\)'

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Q.31

'一个被4条直线围成的长方形,是由2条垂直线和2条水平线组合而成的,所以所求个数为${}_5 C_2 \\times {}_5 C_2={\\left(\\frac{5 \\cdot 4}{2 \\cdot 1}\\right)}^2=10^2=100 \\text{(个)}'

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Q.32

'使用0、1、2、3、4、5这6种数字,能够组成多少个不超过4位的正整数?允许重复使用同一个数字。'

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Q.33

'A市和B市之间有5条独立的公交线路。在以下情况下,有多少种方法可以往返A市和B市。'

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Q.34

'假设有4颗白色珠子,3颗黑色珠子,1颗红色珠子。将它们排成一行有\ \\square \种方法,排成一个圆有\ \\square \种方法。此外,通过这些珠子穿线,制作成环的方法有\ \\square \种。'

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Q.35

'请分解以下方程。'

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Q.36

'请因式分解以下表达式。'

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Q.37

'练习答案 1 (1) \ -x^{2}+5 x-1 \ (2) \ -3 x^{2}+3 x y-4 y^{2} \'

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Q.38

'(4)尽量因式分解3x3+(9y+z)x23y(z+2y)x+2y2z -3x^{3} + (9y + z)x^{2} - 3y(z + 2y)x + 2y^{2}z 。'

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Q.39

'请计算以下表达式。'

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Q.40

'(2) 2 x^{2}-4 x+2=2(x^{2}-2 x+1)'

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Q.41

'請因數分解以下方程式。'

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Q.42

'请因式分解以下表达式。'

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Q.43

'数,字母和它们的乘积表达式称为什么?'

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Q.44

'二次展开公式'

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Q.45

''

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Q.46

'展开下列表达式。'

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Q.47

'展开以下表达式。'

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Q.48

''

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Q.49

'整理以下方程式,按照x(1),x(2),a(3)的次幂顺序。'

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Q.50

'整理以下表达式,按照x的降幂顺序。'

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Q.51

'请将以下二次方程完成平方。'

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Q.52

'请因式分解以下表达式。'

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Q.53

'给定多项式P=3x^{3}-3xy^{2}+x^{2}-y^{2}+ax+by。'

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Q.54

'展开公式是 (a^2) - (b^2)'

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Q.55

'请计算以下表达式。'

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Q.56

'使用因式分解公式展开以下表达式。'

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Q.57

'(a-b)^{2}的展开公式是a^{2}-2ab+b^{2}'

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Q.58

''

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Q.59

'请分解以下表达式。'

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Q.60

'在第2节“多项式的乘法”中,我们学习了如何展开多项式的乘积形式,并将其表示为一个多项式的方法。现在,我们将学习相反的过程,即将一个多项式表示为单项式或多项式的乘积形式。'

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Q.61

'展开以下表达式。'

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Q.62

'(1) \7 x^{2} + 4 x - 17\ (2) \\(x^{2}-(2 a-b) x-a\\) (3) \\(-a^{2}-2(7 b-2) a+2 b^{2}+2 b-5\\)'

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Q.63

'展开以下表达式:x(x-1)(x+1)(x^2+1)(x^4+1)'

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Q.64

'将函数y=f(x)的图像相对于原点对称移动时所代表的函数是y=-f(-x)。如果a和b是实数,并且函数f(x)=x^{2}+ax+b在0 <= x <= 1处的最小值为m,则用a和b来表示m。'

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Q.65

'(7) \ 4 x^{4}-13 x^{2} y^{2}+9 y^{4} \'

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Q.66

'将给定方程y=-x^{2}+2 x进行转换'

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Q.67

'请分解以下表达式。'

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Q.68

'展开以下表达式:\n(x+2y)^2(x^2+4y^2)^2(x-2y)^2'

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Q.69

'请使用乘法计算以下多项式:(x + 2)(x - 3)'

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Q.70

'请因式分解以下多项式。'

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Q.71

'请因式分解以下表达式。'

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Q.72

'确定单项式的次数和系数。还可以确定方括号内的字母对应的次数和系数。'

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Q.73

'请完成以下二次方程的平方'

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Q.74

'请因式分解以下表达式。'

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Q.75

'计算以下表达式:(4) (√3 + √5)²'

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Q.76

'从7名成员中选择部长、副部长和财务的方法有多少种?请注意,不允许兼任。'

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Q.77

'请因数分解以下表达式。'

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Q.78

'展开以下表达式。'

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Q.79

'与以往不同,考虑允许重复使用相同物品的排列。例如,如果从两种字符A和B中允许重复地取3个字符排成一行,则树状图如右所示,总共有2^{3} (种方式)。'

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Q.80

'求下列方程的值。'

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Q.81

''

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Q.82

'请因式分解以下表达式。'

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Q.83

''

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Q.84

'(1) 展开以下表达式。(2) (3 x-1)^{3}(3) (3 x^{2}-a)(9 x^{4}+3 a x^{2}+a^{2})(4) (x-1)(x+1)(x^{2}+x+1)(x^{2}-x+1)(5) (x+2)(x+4)(x-3)(x-5)(6) (x+1)^{3}(x-1)^{3}'

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Q.85

'请因式分解以下表达式。'

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Q.86

''

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Q.87

'分解下列方程式。(1) 8x³+1 (2) 64a³-125b³'

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Q.88

''

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Q.89

'(ア) -x^{2}+8 x\n(1) -x^{2}+16\n(ウ) 2\n(I) 24'

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Q.90

'请因式分解以下表达式。(1) x^3 + 2x^2y - x^2z + xy^2 - 2xyz - y^2z (2) x^3 + 3x^2y + zx^2 + 2xy^2 + 3xyz + 2zy^2'

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Q.91

'将给定表达式进行转换,并求出最大值和最小值: (1) 3x^2 + 4y^2 进行转换并代入。 (2) 根据x和y的范围求最大值和最小值。 (3) 当x为实数时,y = (x^2 + 2x)^2 + 8(x^2 + 2x) + 10 进行转换,并令t = x^2 + 2x。求最大值和最小值。'

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Q.92

'在展开的表达式中,x^5的系数为甲,x^3的系数为乙。'

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Q.93

'将 10 名学生分成一些小组。这时有多少种将其分为(1) 2 人,3 人,5 人的 3 个小组的方法。(2) 3 人,3 人,4 人的 3 个小组的方法。(3) 2 人,2 人,3 人,3 人的 4 个小组的方法。'

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Q.94

''

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Q.95

'将抛物线y=x^{2}+a x+b关于原点对称移动后的抛物线方程是,将x,y分别替换为-x,-y得到-y=(- x)^{2}+a(-x)+b即y=-x^{2}+a x-b。将抛物线y=-x^{2}+a x-b沿x轴方向移动3个单位,y轴方向移动6个单位,新抛物线方程是y-6=-(x-3)^{2}+a(x-3)-b即y=-x^{2}+(a+6) x-3a-b-3,这等于y=-x^{2}+4 x-7,因此a+6=4,-3a-b-3=-7,解得a=-2,b=10。'

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Q.96

'展开以下表达式。'

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Q.97

'包含相同物品的排列'

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Q.98

'请因式分解表达式(3)(x + 1)(x + 2)(x + 3)(x + 4) - 3。'

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Q.99

'请分解以下表达式。'

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Q.00

'(9) 将下列表达式因式分解:(ab)x2+(ba)xy (a-b) x^{2}+(b-a) x y '

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Q.01

'请将以下表达式进行因式分解。'

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Q.02

'展开(a+b+c)(x+y)(p+q)后,会得到多少项?'

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Q.03

'将抛物线y=ax^{2}+bx+c沿x轴方向平行移动2个单位,y轴方向平行移动-1个单位,得到抛物线33y=-2x^{2}+3。求解系数a、b、c的值。'

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Q.04

'(a+b)^{2}的展开公式是:a^{2} + 2ab + b^{2}'

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Q.05

'展开以下表达式。'

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Q.06

'请因式分解以下表达式:(1) 6x^{2}+13x+6 (2) 3a^{2}-11a+6 (3) 12x^{2}+5x-2 (4) 6x^{2}-5x-4 (5) 4x^{2}-4x-15 (6) 6a^{2}+17ab+12b^{2} (7) 6x^{2}+5xy-21y^{2} (8) 12x^{2}-8xy-15y^{2} (9) 4x^{2}-3xy-27y^{2}'

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Q.07

'从4名学生中选出1名主席和1名副主席时,有多少种选择方式?要求主席和副主席不能兼任。'

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Q.08

'展开以下表达式:(x+1)(x+2)(x-1)(x-2)'

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Q.09

'(1) 展开表达式 (x-2 y+1)(x-2 y-2)'

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Q.10

'如果有3名候选人,并且有10人进行匿名投票,那么有多少种投票方式?'

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Q.11

''

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Q.12

'1. (1)次数3,系数为2;aa:次数1,系数为2x22x^{2}\n2. 次数17,系数为\x0crac13-\x0crac{1}{3}yy:次数7,系数为\x0crac13ab7x2-\x0crac{1}{3}ab^{7}x^{2}aabb:次数8,系数为\x0crac13x2y7-\x0crac{1}{3}x^{2}y^{7}'

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Q.13

'(3) x^{3}+2 x^{2}-9 x-18\nx^{3}+2 x^{2}-9 x-18=(x^{3}+2 x^{2})-(9 x+18)=x^{2}(x+2)-9(x+2)=(x+2)…'

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Q.14

'展开以下表达式。'

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Q.15

'请计算一种,三种和四种香料的图案数量,并找出每种情况的可能性。'

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