第43页

信息发布者：
不变
相加减
$\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}$
通分
$\frac{a}{b} \pm \frac{c}{d} = \frac{ad}{bd} \pm \frac{bc}{bd} = \frac{ad \pm bc}{bd}$

D
A
$\frac{3y + 2x}{6x^{2}y^{2}}$
$\frac{a - 2}{a}$
解：
$\begin{aligned}&\frac{5a + 3b}{a^{2}-b^{2}}-\frac{2a}{a^{2}-b^{2}}\\=&\frac{5a + 3b - 2a}{a^{2}-b^{2}}\\=&\frac{3a + 3b}{(a + b)(a - b)}\\=&\frac{3(a + b)}{(a + b)(a - b)}\\=&\frac{3}{a - b}\end{aligned}$
解：
$\begin{aligned}&\frac{a}{a - 1}-\frac{3a - 1}{a^{2}-1}\\=&\frac{a}{a - 1}-\frac{3a - 1}{(a + 1)(a - 1)}\\=&\frac{a(a + 1)}{(a + 1)(a - 1)}-\frac{3a - 1}{(a + 1)(a - 1)}\\=&\frac{a^{2}+a-(3a - 1)}{(a + 1)(a - 1)}\\=&\frac{a^{2}+a - 3a + 1}{(a + 1)(a - 1)}\\=&\frac{a^{2}-2a + 1}{(a + 1)(a - 1)}\\=&\frac{(a - 1)^{2}}{(a + 1)(a - 1)}\\=&\frac{a - 1}{a + 1}\end{aligned}$
解：
$\begin{aligned}&m + 2-\frac{m^{2}+4}{m - 2}\\=&\frac{(m + 2)(m - 2)}{m - 2}-\frac{m^{2}+4}{m - 2}\\=&\frac{m^{2}-4-(m^{2}+4)}{m - 2}\\=&\frac{m^{2}-4 - m^{2}-4}{m - 2}\\=&\frac{-8}{m - 2}\end{aligned}$
解：
$\begin{aligned}&a - b-\frac{(a + b)^{2}}{a + b}\\=&a - b-(a + b)\\=&a - b - a - b\\=&-2b\end{aligned}$
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