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【典型例题1】计算：
(1)$\dfrac{a^{2}+b}{2ab}-\dfrac{a^{2}-b}{2ab}$；
(2)$\dfrac{3a}{b - a}-\dfrac{b}{b - a}-\dfrac{a + b}{b - a}$.
【解】(1)原式$=\dfrac{a^{2}+b - a^{2}+b}{2ab}= \dfrac{2b}{2ab}= \dfrac{1}{a}$.
(2)原式$=\dfrac{3a - b - a - b}{b - a}= \dfrac{-2(b - a)}{b - a}= -2$.
(1)$\dfrac{a^{2}+b}{2ab}-\dfrac{a^{2}-b}{2ab}$；
(2)$\dfrac{3a}{b - a}-\dfrac{b}{b - a}-\dfrac{a + b}{b - a}$.
【解】(1)原式$=\dfrac{a^{2}+b - a^{2}+b}{2ab}= \dfrac{2b}{2ab}= \dfrac{1}{a}$.
(2)原式$=\dfrac{3a - b - a - b}{b - a}= \dfrac{-2(b - a)}{b - a}= -2$.
答案
(1)
原式$=\dfrac{a^{2}+b}{2ab}-\dfrac{a^{2}-b}{2ab}$
$=\dfrac{a^{2}+b - (a^{2}-b)}{2ab}$
$=\dfrac{a^{2}+b - a^{2}+b}{2ab}$
$=\dfrac{2b}{2ab}$
$=\dfrac{1}{a}$
(2)
原式$=\dfrac{3a}{b - a}-\dfrac{b}{b - a}-\dfrac{a + b}{b - a}$
$=\dfrac{3a - b-(a + b)}{b - a}$
$=\dfrac{3a - b - a - b}{b - a}$
$=\dfrac{2a-2b}{b - a}$
$=\dfrac{-2(b - a)}{b - a}$
$=-2$
原式$=\dfrac{a^{2}+b}{2ab}-\dfrac{a^{2}-b}{2ab}$
$=\dfrac{a^{2}+b - (a^{2}-b)}{2ab}$
$=\dfrac{a^{2}+b - a^{2}+b}{2ab}$
$=\dfrac{2b}{2ab}$
$=\dfrac{1}{a}$
(2)
原式$=\dfrac{3a}{b - a}-\dfrac{b}{b - a}-\dfrac{a + b}{b - a}$
$=\dfrac{3a - b-(a + b)}{b - a}$
$=\dfrac{3a - b - a - b}{b - a}$
$=\dfrac{2a-2b}{b - a}$
$=\dfrac{-2(b - a)}{b - a}$
$=-2$
1. 计算：
(1)$\dfrac{3a^{2}-5a}{a^{2}+1}+\dfrac{2a^{2}+4}{a^{2}+1}-\dfrac{2a^{2}-5a + 1}{a^{2}+1}$；
(2)$\dfrac{a - 3b}{a - b}+\dfrac{a + b}{a - b}$.
(1)$\dfrac{3a^{2}-5a}{a^{2}+1}+\dfrac{2a^{2}+4}{a^{2}+1}-\dfrac{2a^{2}-5a + 1}{a^{2}+1}$；
(2)$\dfrac{a - 3b}{a - b}+\dfrac{a + b}{a - b}$.
答案
(1)
$\begin{aligned}&\dfrac{3a^{2}-5a}{a^{2}+1}+\dfrac{2a^{2}+4}{a^{2}+1}-\dfrac{2a^{2}-5a + 1}{a^{2}+1}\\=&\dfrac{(3a^{2}-5a)+(2a^{2}+4)-(2a^{2}-5a + 1)}{a^{2}+1}\\=&\dfrac{3a^{2}-5a + 2a^{2}+4 - 2a^{2}+5a - 1}{a^{2}+1}\\=&\dfrac{(3a^{2}+2a^{2}-2a^{2}) + (-5a + 5a) + (4 - 1)}{a^{2}+1}\\=&\dfrac{3a^{2} + 3}{a^{2}+1}\\=&\dfrac{3(a^{2}+1)}{a^{2}+1}\\=&3\end{aligned}$
(2)
$\begin{aligned}&\dfrac{a - 3b}{a - b}+\dfrac{a + b}{a - b}\\=&\dfrac{(a - 3b)+(a + b)}{a - b}\\=&\dfrac{a - 3b + a + b}{a - b}\\=&\dfrac{2a - 2b}{a - b}\\=&\dfrac{2(a - b)}{a - b}\\=&2\end{aligned}$
$\begin{aligned}&\dfrac{3a^{2}-5a}{a^{2}+1}+\dfrac{2a^{2}+4}{a^{2}+1}-\dfrac{2a^{2}-5a + 1}{a^{2}+1}\\=&\dfrac{(3a^{2}-5a)+(2a^{2}+4)-(2a^{2}-5a + 1)}{a^{2}+1}\\=&\dfrac{3a^{2}-5a + 2a^{2}+4 - 2a^{2}+5a - 1}{a^{2}+1}\\=&\dfrac{(3a^{2}+2a^{2}-2a^{2}) + (-5a + 5a) + (4 - 1)}{a^{2}+1}\\=&\dfrac{3a^{2} + 3}{a^{2}+1}\\=&\dfrac{3(a^{2}+1)}{a^{2}+1}\\=&3\end{aligned}$
(2)
$\begin{aligned}&\dfrac{a - 3b}{a - b}+\dfrac{a + b}{a - b}\\=&\dfrac{(a - 3b)+(a + b)}{a - b}\\=&\dfrac{a - 3b + a + b}{a - b}\\=&\dfrac{2a - 2b}{a - b}\\=&\dfrac{2(a - b)}{a - b}\\=&2\end{aligned}$
【典型例题2】计算：(1)$\dfrac{7a}{8b^{2}c}-\dfrac{b}{12a^{2}c}$；
(2)$\dfrac{24}{x^{2}-16}+\dfrac{3}{4 - x}$.
思路导引 异分母的分式相加减,先找出最简公分母进行通分,变为同分母的分式相加减,最后结果一定要化成最简分式或整式.
【解】(1)原式$=\dfrac{21a^{3}}{24a^{2}b^{2}c}-\dfrac{2b^{3}}{24a^{2}b^{2}c}= \dfrac{21a^{3}-2b^{3}}{24a^{2}b^{2}c}$.
(2)原式$=\dfrac{24}{(x + 4)(x - 4)}-\dfrac{3(x + 4)}{(x + 4)(x - 4)}= \dfrac{24 - 3(x + 4)}{(x + 4)(x - 4)}= \dfrac{24 - 3x - 12}{(x + 4)(x - 4)}= \dfrac{-3x + 12}{(x + 4)(x - 4)}= \dfrac{-3(x - 4)}{(x + 4)(x - 4)}= -\dfrac{3}{x + 4}$.
(2)$\dfrac{24}{x^{2}-16}+\dfrac{3}{4 - x}$.
思路导引 异分母的分式相加减,先找出最简公分母进行通分,变为同分母的分式相加减,最后结果一定要化成最简分式或整式.
【解】(1)原式$=\dfrac{21a^{3}}{24a^{2}b^{2}c}-\dfrac{2b^{3}}{24a^{2}b^{2}c}= \dfrac{21a^{3}-2b^{3}}{24a^{2}b^{2}c}$.
(2)原式$=\dfrac{24}{(x + 4)(x - 4)}-\dfrac{3(x + 4)}{(x + 4)(x - 4)}= \dfrac{24 - 3(x + 4)}{(x + 4)(x - 4)}= \dfrac{24 - 3x - 12}{(x + 4)(x - 4)}= \dfrac{-3x + 12}{(x + 4)(x - 4)}= \dfrac{-3(x - 4)}{(x + 4)(x - 4)}= -\dfrac{3}{x + 4}$.
答案
(1)原式$=\dfrac{7a \cdot 3a^{2}}{8b^{2}c \cdot 3a^{2}} - \dfrac{b \cdot 2b^{2}}{12a^{2}c \cdot 2b^{2}} = \dfrac{21a^{3}}{24a^{2}b^{2}c} - \dfrac{2b^{3}}{24a^{2}b^{2}c} = \dfrac{21a^{3} - 2b^{3}}{24a^{2}b^{2}c}$
(2)原式$=\dfrac{24}{(x + 4)(x - 4)} + \dfrac{3}{-(x - 4)} = \dfrac{24}{(x + 4)(x - 4)} - \dfrac{3}{x - 4} = \dfrac{24}{(x + 4)(x - 4)} - \dfrac{3(x + 4)}{(x + 4)(x - 4)} = \dfrac{24 - 3(x + 4)}{(x + 4)(x - 4)} = \dfrac{24 - 3x - 12}{(x + 4)(x - 4)} = \dfrac{-3x + 12}{(x + 4)(x - 4)} = \dfrac{-3(x - 4)}{(x + 4)(x - 4)} = -\dfrac{3}{x + 4}$
(2)原式$=\dfrac{24}{(x + 4)(x - 4)} + \dfrac{3}{-(x - 4)} = \dfrac{24}{(x + 4)(x - 4)} - \dfrac{3}{x - 4} = \dfrac{24}{(x + 4)(x - 4)} - \dfrac{3(x + 4)}{(x + 4)(x - 4)} = \dfrac{24 - 3(x + 4)}{(x + 4)(x - 4)} = \dfrac{24 - 3x - 12}{(x + 4)(x - 4)} = \dfrac{-3x + 12}{(x + 4)(x - 4)} = \dfrac{-3(x - 4)}{(x + 4)(x - 4)} = -\dfrac{3}{x + 4}$
2. 计算：
(1)$\dfrac{2c}{b^{2}-c^{2}}-\dfrac{1}{b + c}+\dfrac{1}{c - b}$；
(2)$x - y+\dfrac{2y^{2}}{x + y}$.
(1)$\dfrac{2c}{b^{2}-c^{2}}-\dfrac{1}{b + c}+\dfrac{1}{c - b}$；
(2)$x - y+\dfrac{2y^{2}}{x + y}$.
答案
(1)
首先，对分母进行因式分解，$b^{2}-c^{2}=(b + c)(b - c)$，$c - b=-(b - c)$。
原式$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{1}{b + c}-\dfrac{1}{b - c}$
通分，通分后的分母为$(b + c)(b - c)$。
$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{b - c}{(b + c)(b - c)}-\dfrac{b + c}{(b + c)(b - c)}$
$=\dfrac{2c-(b - c)-(b + c)}{(b + c)(b - c)}$
$=\dfrac{2c - b + c - b - c}{(b + c)(b - c)}$
$=\dfrac{2c - 2b}{(b + c)(b - c)}$
$=\dfrac{2(c - b)}{(b + c)(b - c)}$
$=-\dfrac{2}{b + c}$
(2)
先对式子中的$x - y$通分，通分后的分母为$x + y$。
原式$=\dfrac{(x - y)(x + y)}{x + y}+\dfrac{2y^{2}}{x + y}$
$=\dfrac{x^{2}-y^{2}+2y^{2}}{x + y}$
$=\dfrac{x^{2}+y^{2}}{x + y}$
综上，(1)的答案是$-\dfrac{2}{b + c}$；(2)的答案是$\dfrac{x^{2}+y^{2}}{x + y}$。
首先，对分母进行因式分解，$b^{2}-c^{2}=(b + c)(b - c)$，$c - b=-(b - c)$。
原式$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{1}{b + c}-\dfrac{1}{b - c}$
通分，通分后的分母为$(b + c)(b - c)$。
$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{b - c}{(b + c)(b - c)}-\dfrac{b + c}{(b + c)(b - c)}$
$=\dfrac{2c-(b - c)-(b + c)}{(b + c)(b - c)}$
$=\dfrac{2c - b + c - b - c}{(b + c)(b - c)}$
$=\dfrac{2c - 2b}{(b + c)(b - c)}$
$=\dfrac{2(c - b)}{(b + c)(b - c)}$
$=-\dfrac{2}{b + c}$
(2)
先对式子中的$x - y$通分，通分后的分母为$x + y$。
原式$=\dfrac{(x - y)(x + y)}{x + y}+\dfrac{2y^{2}}{x + y}$
$=\dfrac{x^{2}-y^{2}+2y^{2}}{x + y}$
$=\dfrac{x^{2}+y^{2}}{x + y}$
综上，(1)的答案是$-\dfrac{2}{b + c}$；(2)的答案是$\dfrac{x^{2}+y^{2}}{x + y}$。
