7. 化简$x^{3}·(\frac{y^{3}}{x})^{2}$的结果是 (
A
)
A.$xy^{6}$
B.$xy^{5}$
C.$x^{2}y^{5}$
D.$x^{2}y^{6}$
答案：7. A
解析：
$x^{3}·(\frac{y^{3}}{x})^{2}=x^{3}·\frac{y^{6}}{x^{2}}=xy^{6}$，结果为A。
8. 化简：
(1) (2025·安徽)$\frac{2}{x^{2}+2x + 1}÷\frac{1}{x^{2}-1}=$
$\frac{2x - 2}{x + 1}$
；
(2) $\frac{2}{m^{2}-m}÷\frac{m}{1 + m^{2}-2m}=$
$\frac{2m - 2}{m^{2}}$
.
答案：8. (1) $\frac{2x - 2}{x + 1}$ (2) $\frac{2m - 2}{m^{2}}$
解析：
(1) $\frac{2}{x^{2}+2x + 1}÷\frac{1}{x^{2}-1}$
$=\frac{2}{(x+1)^2}×(x^2 - 1)$
$=\frac{2}{(x+1)^2}×(x + 1)(x - 1)$
$=\frac{2(x - 1)}{x + 1}$
$=\frac{2x - 2}{x + 1}$
(2) $\frac{2}{m^{2}-m}÷\frac{m}{1 + m^{2}-2m}$
$=\frac{2}{m(m - 1)}÷\frac{m}{(m - 1)^2}$
$=\frac{2}{m(m - 1)}×\frac{(m - 1)^2}{m}$
$=\frac{2(m - 1)}{m^2}$
$=\frac{2m - 2}{m^2}$
9. (教材变式)计算：
(1) $\frac{3y(x + 3)}{x - 3}·\frac{2(x - 3)}{9y^{2}}$；
(2) $\frac{x^{4}-y^{4}}{x^{2}-2xy + y^{2}}÷\frac{x^{2}+y^{2}}{y - x}$.
答案：9. (1) $\frac{2(x + 3)}{3y}$ (2) $-x - y$
解析：
(1) $\frac{3y(x + 3)}{x - 3}·\frac{2(x - 3)}{9y^{2}}$
$=\frac{3y(x + 3)·2(x - 3)}{(x - 3)·9y^{2}}$
$=\frac{6y(x + 3)(x - 3)}{9y^{2}(x - 3)}$
$=\frac{2(x + 3)}{3y}$
(2) $\frac{x^{4}-y^{4}}{x^{2}-2xy + y^{2}}÷\frac{x^{2}+y^{2}}{y - x}$
$=\frac{(x^{2}+y^{2})(x^{2}-y^{2})}{(x - y)^{2}}·\frac{y - x}{x^{2}+y^{2}}$
$=\frac{(x^{2}+y^{2})(x + y)(x - y)}{(x - y)^{2}}·\frac{-(x - y)}{x^{2}+y^{2}}$
$=-(x + y)$
$=-x - y$
10. 先化简,再求值：$\frac{y(x - y)-x(x + y)}{x^{2}-y^{2}}÷\frac{x^{2}+y^{2}}{x + y}$,其中$x = 2,y = -1$.
答案：10. 原式 = $-\frac{1}{x - y}$. 当 $x = 2, y = -1$ 时, 原式 = $-\frac{1}{3}$
解析：
解：原式$=\frac{xy - y^{2} - x^{2} - xy}{(x + y)(x - y)} · \frac{x + y}{x^{2} + y^{2}}$
$=\frac{-x^{2} - y^{2}}{(x + y)(x - y)} · \frac{x + y}{x^{2} + y^{2}}$
$=\frac{-(x^{2} + y^{2})}{(x + y)(x - y)} · \frac{x + y}{x^{2} + y^{2}}$
$=-\frac{1}{x - y}$
当$x = 2, y = -1$时，原式$=-\frac{1}{2 - (-1)}=-\frac{1}{3}$
11. 已知实数$x,y$满足$\sqrt{x - 3}+y^{2}-4y + 4 = 0$,求代数式$\frac{x^{2}-y^{2}}{xy}·\frac{1}{x^{2}-2xy + y^{2}}÷\frac{x}{x^{2}y - xy^{2}}$的值.
答案：11. 原式 = $\frac{x + y}{x}$
∵ $\sqrt{x - 3} + y^{2} - 4y + 4 = 0$,
∴ $\sqrt{x - 3} + (y - 2)^{2} = 0$. 根据非负数的性质, 得 $x = 3, y = 2$.
∴ 原式 = $\frac{3 + 2}{3} = \frac{5}{3}$
