2025年通成学典课时作业本八年级数学下册苏科版苏州专版第79页答案
7. (2023·河北)化简$x^{3}\cdot(\frac{y^{3}}{x})^{2}$的结果是 ( )
A. $xy^{6}$
B. $xy^{5}$
C. $x^{2}y^{5}$
D. $x^{2}y^{6}$

答案

A
8. 化简:
(1)$\frac{a^{2}+2a + 1}{a^{2}-1}\cdot\frac{1}{a + 1}=$__________;
(2)$\frac{2}{m^{2}-m}\div\frac{m}{1 + m^{2}-2m}=$__________.

答案

(1) $\frac{1}{a - 1}$ (2) $\frac{2m - 2}{m^{2}}$
9. 计算:
(1)$\frac{3y(x + 3)}{x - 3}\cdot\frac{2(x - 3)}{9y^{2}}$;
(2)$\frac{x^{4}-y^{4}}{x^{2}-2xy + y^{2}}\div\frac{x^{2}+y^{2}}{y - x}$.

答案

(1) $\frac{2(x + 3)}{3y}$ (2) $-x - y$
10. 先化简,再求值:$\frac{y(x - y)-x(x + y)}{x^{2}-y^{2}}\div\frac{x^{2}+y^{2}}{x + y}$,其中$x = 2$,$y = -1$.

答案

原式$=-\frac{1}{x - y}$. 当$x = 2$, $y = -1$时, 原式$=-\frac{1}{3}$
11. 已知实数$x、y$满足$\sqrt{x - 3}+y^{2}-4y + 4 = 0$,求代数式$\frac{x^{2}-y^{2}}{xy}\cdot\frac{1}{x^{2}-2xy + y^{2}}\div\frac{x}{x^{2}y - xy^{2}}$的值.

答案

原式$=\frac{x + y}{x}$. $\because \sqrt{x - 3} + y^{2} - 4y + 4 = 0$,
$\therefore \sqrt{x - 3} + (y - 2)^{2} = 0$. 根据非负数的性质, 得$x = 3$,
$y = 2$. $\therefore$ 原式$=\frac{3 + 2}{3}=\frac{5}{3}$