2026年学法大视野八年级数学上册人教版第66页答案
【例】计算:
(1)$3a^2b · (ab - 4b^2)$;
(2)$( \frac{3}{4}x^2y - \frac{1}{2}xy^2 - \frac{5}{6}y^2 )(-4xy^2)$;
(3)$(-3xy^2)^2(4x - y + 1)$;
(4)$x(x - 1) + 2x(x + 1) - 3x(2x - 5)$。

答案

(1) 原式=$3a^3b^2 - 12a^2b^3$.
(2) 原式=$-3x^3y^3 + 2x^2y^4 + \frac{10}{3}xy^4$.
(3) 原式=$9x^2y^4 · (4x - y + 1)=36x^3y^4 - 9x^2y^5 + 9x^2y^4$.
(4) 原式=$(x^2 - x)+(2x^2 + 2x)-(6x^2 - 15x)$
$=x^2 - x + 2x^2 + 2x - 6x^2 + 15x$
$=-3x^2 + 16x$.
【变式1】下列计算错误的是 (
C
)

A.$-3x(2 - x) = -6x + 3x^2$
B.$(2m^2n - 3mn^2)(-mn) = -2m^3n^2 + 3m^2n^3$
C.$xy(x^2y - 3xy^2 - 1) = x^3y^2 - x^2y^3$
D.$(\frac{2}{5}x^{n+1} - \frac{1}{3}y)xy = \frac{2}{5}x^{n+2}y - \frac{1}{3}xy^2$

答案

C
【变式2】先化简,再求值:$(-3a)^2 - 2a(-ab + 3b^2) + 4(ab^2 - \frac{1}{2}a^2b - \frac{9}{4}a^2)$,其中$a,b$满足$(a-4)^2 + \left|b + \frac{3}{2}\right| = 0$。

答案

解:$(-3a)^2 - 2a(-ab + 3b^2) + 4(ab^2 - \frac{1}{2}a^2b - \frac{9}{4}a^2)$
$=9a^2 + 2a^2b - 6ab^2 + 4ab^2 - 2a^2b - 9a^2 = -2ab^2$.
$\because (a-4)^2 + \left|b + \frac{3}{2}\right| = 0,(a-4)^2≥0,\left|b + \frac{3}{2}\right|≥0,$
$\therefore (a-4)^2=0,\left|b + \frac{3}{2}\right|=0,\therefore a-4=0,b + \frac{3}{2}=0,$
$\therefore a=4,b=-\frac{3}{2},$
$\therefore 原式=-2×4×(-\frac{3}{2})^2=-8×\frac{9}{4}=-18.$
1. 计算:$3a(a^2b^3 + 2ab^2) = (\quad)$

A.$3a^2b^3 + 2ab^2$
B.$3a^3b^3 + 6ab^2$
C.$3a^3b^3 + 2ab^2$
D.$3a^3b^3 + 6a^2b^2$

答案

D
2. 计算$(-3x)(-2x^2+\frac{2}{3}x-4)$的结果是(
D
)

A.$-6x^3-2x^2+12x$
B.$6x^3-2x^2+12$
C.$6x^3+2x^2-12x$
D.$6x^3-2x^2+12x$

答案

2. D
3. 计算$(-m^2)^3 · (2m+1)$的结果是(
A
)

A.$-2m^7 - m^6$
B.$-2m^6 + m^6$
C.$-2m^7 - m^5$
D.$-2m^6 - m^5$

答案

3. A