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题型3 先化简,再求值：$6a^{2}-5a(-a + 2b - 1)+4a(-3a-\frac{5}{2}b-\frac{3}{4})$,其中$a = 2$,$b= \frac{1}{20}$.
答案
解:原式=6a²+5a²-10ab+5a-12a²-10ab-3a=-a²-20ab+2a. 当a=2,b=1/20时,原式=-2²-20×2×1/20+2×2=-2.
解析
解:原式$=6a^{2}+5a^{2}-10ab+5a-12a^{2}-10ab-3a=-a^{2}-20ab+2a$.
当$a = 2$,$b=\frac{1}{20}$时,原式$=-2^{2}-20×2×\frac{1}{20}+2×2=-4 - 2 + 4=-2$.
当$a = 2$,$b=\frac{1}{20}$时,原式$=-2^{2}-20×2×\frac{1}{20}+2×2=-4 - 2 + 4=-2$.
题型4 已知$9a^{n}b与-2a^{m + 1}b^{2n}的积与5a^{4}b^{3}$是同类项,求$m$,$n$的值.
答案
解:9aⁿb·(-2aᵐ⁺¹b²ⁿ)=9×(-2)×(aⁿ·aᵐ⁺¹)×(b·b²ⁿ)=-18aⁿ⁺ᵐ⁺¹b²ⁿ⁺¹. 因为-18aⁿ⁺ᵐ⁺¹b²ⁿ⁺¹与5a⁴b³是同类项,所以{n+m+1=4,2n+1=3,解得{m=2,n=1,即m=2,n=1.
解析
解：$9a^{n}b \cdot (-2a^{m + 1}b^{2n})$
$=9×(-2)×(a^{n}\cdot a^{m + 1})×(b\cdot b^{2n})$
$=-18a^{n + m + 1}b^{2n + 1}$
因为$-18a^{n + m + 1}b^{2n + 1}$与$5a^{4}b^{3}$是同类项，所以$\begin{cases}n + m + 1 = 4 \\2n + 1 = 3\end{cases}$
由$2n + 1 = 3$，得$2n = 2$，$n = 1$
将$n = 1$代入$n + m + 1 = 4$，得$1 + m + 1 = 4$，$m = 2$
即$m = 2$，$n = 1$
$=9×(-2)×(a^{n}\cdot a^{m + 1})×(b\cdot b^{2n})$
$=-18a^{n + m + 1}b^{2n + 1}$
因为$-18a^{n + m + 1}b^{2n + 1}$与$5a^{4}b^{3}$是同类项，所以$\begin{cases}n + m + 1 = 4 \\2n + 1 = 3\end{cases}$
由$2n + 1 = 3$，得$2n = 2$，$n = 1$
将$n = 1$代入$n + m + 1 = 4$，得$1 + m + 1 = 4$，$m = 2$
即$m = 2$，$n = 1$
1. 计算$(-3x)\cdot (2x^{2}-5x - 1)$的结果是(
A.$-6x^{2}-15x^{2}-3x$
B.$-6x^{3}+15x^{2}+3x$
C.$-6x^{3}+15x^{2}$
D.$-6x^{3}+15x^{2}-1$
B
)A.$-6x^{2}-15x^{2}-3x$
B.$-6x^{3}+15x^{2}+3x$
C.$-6x^{3}+15x^{2}$
D.$-6x^{3}+15x^{2}-1$
答案
B
解析
$(-3x)\cdot(2x^{2}-5x - 1)$
$=(-3x)\cdot2x^{2}+(-3x)\cdot(-5x)+(-3x)\cdot(-1)$
$=-6x^{3}+15x^{2}+3x$
B
$=(-3x)\cdot2x^{2}+(-3x)\cdot(-5x)+(-3x)\cdot(-1)$
$=-6x^{3}+15x^{2}+3x$
B
2. 已知$ab^{2}= -2$,则$-ab(a^{2}b^{5}-ab^{3}+b)$的值为(
A.4
B.2
C.0
D.14
D
)A.4
B.2
C.0
D.14
答案
D
解析
$-ab(a^{2}b^{5}-ab^{3}+b)$
$=-a^{3}b^{6}+a^{2}b^{4}-ab^{2}$
$=-(ab^{2})^{3}+(ab^{2})^{2}-ab^{2}$
因为$ab^{2}=-2$，
所以原式$=-(-2)^{3}+(-2)^{2}-(-2)$
$=-(-8)+4+2$
$=8+4+2$
$=14$
D
$=-a^{3}b^{6}+a^{2}b^{4}-ab^{2}$
$=-(ab^{2})^{3}+(ab^{2})^{2}-ab^{2}$
因为$ab^{2}=-2$，
所以原式$=-(-2)^{3}+(-2)^{2}-(-2)$
$=-(-8)+4+2$
$=8+4+2$
$=14$
D
3. 计算：$(-2a)^{3}(-3a)^{2}= $
-72a⁵
.答案
-72a⁵
解析
$(-2a)^{3}(-3a)^{2}=(-8a^{3})(9a^{2})=-72a^{5}$
4. 计算：
(1)$\frac{1}{2}ab^{2}c\cdot (-0.5ab)^{2}\cdot (-2bc^{2})^{3}$；
(2)$2x^{6}y^{2}\cdot x^{3}y+(-25x^{8}y^{2})(-xy)$.
(1)$\frac{1}{2}ab^{2}c\cdot (-0.5ab)^{2}\cdot (-2bc^{2})^{3}$；
(2)$2x^{6}y^{2}\cdot x^{3}y+(-25x^{8}y^{2})(-xy)$.
答案
解:
(1)原式=1/2ab²c·(1/4a²b²)·(-8b³c⁶)=-a³b⁷c⁷；
(2)原式=2x⁹y³+25x⁹y³=27x⁹y³.
(1)原式=1/2ab²c·(1/4a²b²)·(-8b³c⁶)=-a³b⁷c⁷；
(2)原式=2x⁹y³+25x⁹y³=27x⁹y³.
1. 计算$2x^{2}\cdot (-3x^{3})$的结果是(
A.$-6x^{5}$
B.$6x^{5}$
C.$-2x^{6}$
D.$2x^{6}$
A
)A.$-6x^{5}$
B.$6x^{5}$
C.$-2x^{6}$
D.$2x^{6}$
答案
A
解析
$2x^{2}\cdot (-3x^{3})=[2×(-3)]\cdot(x^{2}\cdot x^{3})=-6x^{5}$，结果为$-6x^{5}$，对应选项A。
A
A
2. 以下计算正确的是(
A.$(-2ab^{2})^{3}= 8a^{3}b^{6}$
B.$3ab + 2b = 5ab$
C.$(-x^{2})\cdot (-2x)^{3}= -8x^{5}$
D.$2m(mn^{2}-3m^{2})= 2m^{2}n^{2}-6m^{3}$
D
)A.$(-2ab^{2})^{3}= 8a^{3}b^{6}$
B.$3ab + 2b = 5ab$
C.$(-x^{2})\cdot (-2x)^{3}= -8x^{5}$
D.$2m(mn^{2}-3m^{2})= 2m^{2}n^{2}-6m^{3}$
答案
D
3. 李老师给同学们出了一道单项式与多项式相乘的题目：$-3x^{2}(2x+□ +1)= -6x^{3}+6x^{2}y - 3x^{2}$,那么“$□$”里应当是(
A.$-y$
B.$-2y$
C.$2y$
D.$2xy$
B
)A.$-y$
B.$-2y$
C.$2y$
D.$2xy$
答案
B
解析
设“$□$”里的单项式为$A$。
$\begin{aligned}-3x^{2}(2x + A + 1)&=-3x^{2}\cdot2x + (-3x^{2})\cdot A + (-3x^{2})\cdot1\\&=-6x^{3} - 3x^{2}A - 3x^{2}\end{aligned}$
已知等式右边为$-6x^{3} + 6x^{2}y - 3x^{2}$，则：
$-3x^{2}A = 6x^{2}y$
解得：
$A = \frac{6x^{2}y}{-3x^{2}} = -2y$
B
$\begin{aligned}-3x^{2}(2x + A + 1)&=-3x^{2}\cdot2x + (-3x^{2})\cdot A + (-3x^{2})\cdot1\\&=-6x^{3} - 3x^{2}A - 3x^{2}\end{aligned}$
已知等式右边为$-6x^{3} + 6x^{2}y - 3x^{2}$，则：
$-3x^{2}A = 6x^{2}y$
解得：
$A = \frac{6x^{2}y}{-3x^{2}} = -2y$
B
