6. 分式$\frac{1}{x^2 - 3x}$与$\frac{1}{x^2 - 9}$通分后的结果是.

$\frac{x+3}{x(x+3)(x-3)},\frac{x}{x(x+3)(x-3)}$

7. 若将分式$\frac{3x^2}{x^2 - y^2}$与分式$\frac{x}{2(x - y)}$通分后，分式$\frac{x}{2(x - y)}$的分母变为$2(x - y)(x + y)$，则分式$\frac{3x^2}{x^2 - y^2}$的分子应变为（）

A.$2(x - y)$
B.$6x^2$
C.$6x^2(x - y)^2$
D.$6x^2(x + y)$

B

8. 如果$\frac{a}{2}=\frac{b}{3}≠0$，那么$\frac{a^2 -4b^2}{a^2 -2ab}$的值是。

4

9. 通分：
(1)$\frac{1}{2x+3}$，$\frac{2}{2x-3}$与$\frac{2x+5}{4x^2 -9}$；
(2)$a-b$，$\frac{b}{a-b}$与$\frac{1}{a^2 -b^2}$。

解: (1) $\dfrac{1}{2x+3}=\dfrac{2x-3}{4x^2-9}$, $\dfrac{2}{2x-3}=\dfrac{2(2x+3)}{4x^2-9}=\dfrac{4x+6}{4x^2-9}$,
$\dfrac{2x+5}{4x^2-9}=\dfrac{2x+5}{4x^2-9}$.
(2) $a - b = \dfrac{(a - b)(a^2 - b^2)}{a^2 - b^2} = \dfrac{(a - b)^2(a + b)}{a^2 - b^2}$, $\dfrac{b}{a - b} = \dfrac{b(a + b)}{a^2 - b^2} = \dfrac{ab + b^2}{a^2 - b^2}$, $\dfrac{1}{a^2 - b^2} = \dfrac{1}{a^2 - b^2}$.

10.已知二元一次方程组$\begin{cases}2x+3y=7, \\3x+2y=-1.\end{cases}$ 求$\frac{x^2 - 2xy + y^2}{x^2 - y^2}$的值.

解:由方程组,得 $x-y=-8,x+y=\dfrac{6}{5}$,
$\therefore$ 原式$=\dfrac{(x-y)^2}{(x+y)(x-y)}=\dfrac{x-y}{x+y}=-\dfrac{20}{3}$.