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7. 化简$x^3 ÷ (\frac{x^3}{y})^2$的结果是(
A.$\frac{x^6}{y^2}$
B.$x^3y^2$
C.$\frac{y^2}{x^3}$
D.$x^2y^6$
C
)A.$\frac{x^6}{y^2}$
B.$x^3y^2$
C.$\frac{y^2}{x^3}$
D.$x^2y^6$
答案
7. C
解析
解：$x^3 ÷ (\frac{x^3}{y})^2$
$=x^3 ÷ \frac{x^6}{y^2}$
$=x^3 × \frac{y^2}{x^6}$
$=\frac{y^2}{x^3}$
C
$=x^3 ÷ \frac{x^6}{y^2}$
$=x^3 × \frac{y^2}{x^6}$
$=\frac{y^2}{x^3}$
C
8. 化简：
(1)$\frac{a^2 + 2a + 1}{a^2 - 1} · \frac{1}{a + 1} =$
(2)$\frac{2}{m^2 - m} ÷ \frac{m}{1 + m^2 - 2m} =$
(1)$\frac{a^2 + 2a + 1}{a^2 - 1} · \frac{1}{a + 1} =$
$\frac{1}{a - 1}$
；(2)$\frac{2}{m^2 - m} ÷ \frac{m}{1 + m^2 - 2m} =$
$\frac{2m - 2}{m^{2}}$
.答案
8. (1) $\frac{1}{a - 1}$ (2) $\frac{2m - 2}{m^{2}}$
解析
(1) $\frac{a^2 + 2a + 1}{a^2 - 1} · \frac{1}{a + 1}$
$=\frac{(a + 1)^2}{(a + 1)(a - 1)} · \frac{1}{a + 1}$
$=\frac{a + 1}{a - 1} · \frac{1}{a + 1}$
$=\frac{1}{a - 1}$
(2) $\frac{2}{m^2 - m} ÷ \frac{m}{1 + m^2 - 2m}$
$=\frac{2}{m(m - 1)} ÷ \frac{m}{(m - 1)^2}$
$=\frac{2}{m(m - 1)} · \frac{(m - 1)^2}{m}$
$=\frac{2(m - 1)}{m^2}$
$=\frac{2m - 2}{m^2}$
$=\frac{(a + 1)^2}{(a + 1)(a - 1)} · \frac{1}{a + 1}$
$=\frac{a + 1}{a - 1} · \frac{1}{a + 1}$
$=\frac{1}{a - 1}$
(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. 计算：
（1）$\frac{3y(x + 3)}{x - 3} · \frac{2(x - 3)}{9y^2}$；
（2）$\frac{x^4 - y^4}{x^2 - 2xy + y^2} ÷ \frac{x^2 + y^2}{y - x}$.
（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) $\frac{x^4 - y^4}{x^2 - 2xy + y^2} ÷ \frac{x^2 + y^2}{y - x}$
$=\frac{(x^2 + y^2)(x^2 - y^2)}{(x - y)^2} · \frac{y - x}{x^2 + y^2}$
$=\frac{(x^2 + y^2)(x + y)(x - y)}{(x - y)^2} · \frac{-(x - y)}{x^2 + y^2}$
$=-(x + y)$
$=-x - y$
$=\frac{3y(x + 3)·2(x - 3)}{(x - 3)·9y^2}$
$=\frac{6y(x + 3)(x - 3)}{9y^2(x - 3)}$
$=\frac{2(x + 3)}{3y}$
(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$=-\frac{1}{2 - (-1)}=-\frac{1}{3}$
$=\frac{-x^2 - y^2}{(x + y)(x - y)} · \frac{x + y}{x^2 + y^2}$
$=-\frac{x^2 + y^2}{(x + y)(x - y)} · \frac{x + y}{x^2 + y^2}$
$=-\frac{1}{x - y}$
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$\frac{x^2 - y^2}{xy} · \frac{1}{x^2 - 2xy + y^2} ÷ \frac{x}{x^2y - xy^2}$的值.
答案
11. 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$=\frac{3 + 2}{3} = \frac{5}{3}$
解析
解：原式$=\frac{(x+y)(x-y)}{xy}·\frac{1}{(x-y)^2}÷\frac{x}{xy(x-y)}$
$=\frac{(x+y)(x-y)}{xy}·\frac{1}{(x-y)^2}·\frac{xy(x-y)}{x}$
$=\frac{x+y}{x}$
$\because\sqrt{x - 3} + y^2 - 4y + 4 = 0$
$\therefore\sqrt{x - 3} + (y - 2)^2 = 0$
$\therefore x - 3 = 0$，$y - 2 = 0$
$\therefore x = 3$，$y = 2$
$\therefore$原式$=\frac{3 + 2}{3}=\frac{5}{3}$
$=\frac{(x+y)(x-y)}{xy}·\frac{1}{(x-y)^2}·\frac{xy(x-y)}{x}$
$=\frac{x+y}{x}$
$\because\sqrt{x - 3} + y^2 - 4y + 4 = 0$
$\therefore\sqrt{x - 3} + (y - 2)^2 = 0$
$\therefore x - 3 = 0$，$y - 2 = 0$
$\therefore x = 3$，$y = 2$
$\therefore$原式$=\frac{3 + 2}{3}=\frac{5}{3}$
