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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)
原式$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{1}{b + c}-\dfrac{1}{b - c}$
$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{b - c}{(b + c)(b - c)}-\dfrac{b + c}{(b + c)(b - c)}$
$=\dfrac{2c-(b - c)-(b + c)}{(b + c)(b - c)}$
$=\dfrac{2c - b + c - b - c}{(b + c)(b - c)}$
$=\dfrac{2c - 2b}{(b + c)(b - c)}$
$=\dfrac{2(c - b)}{(b + c)(b - c)}$
$=-\dfrac{2}{b + c}$
(2)
原式$=\dfrac{(x - y)(x + y)}{x + y}+\dfrac{2y^{2}}{x + y}$
$=\dfrac{x^{2}-y^{2}+2y^{2}}{x + y}$
$=\dfrac{x^{2}+y^{2}}{x + y}$
原式$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{1}{b + c}-\dfrac{1}{b - c}$
$=\dfrac{2c}{(b + c)(b - c)}-\dfrac{b - c}{(b + c)(b - c)}-\dfrac{b + c}{(b + c)(b - c)}$
$=\dfrac{2c-(b - c)-(b + c)}{(b + c)(b - c)}$
$=\dfrac{2c - b + c - b - c}{(b + c)(b - c)}$
$=\dfrac{2c - 2b}{(b + c)(b - c)}$
$=\dfrac{2(c - b)}{(b + c)(b - c)}$
$=-\dfrac{2}{b + c}$
(2)
原式$=\dfrac{(x - y)(x + y)}{x + y}+\dfrac{2y^{2}}{x + y}$
$=\dfrac{x^{2}-y^{2}+2y^{2}}{x + y}$
$=\dfrac{x^{2}+y^{2}}{x + y}$
1. (2024·甘肃中考)计算：$\dfrac{4a}{2a - b}-\dfrac{2b}{2a - b}= $(
A.2
B.$2a - b$
C.$\dfrac{2}{2a - b}$
D.$\dfrac{a - b}{2a - b}$
A
)A.2
B.$2a - b$
C.$\dfrac{2}{2a - b}$
D.$\dfrac{a - b}{2a - b}$
答案
A
解析
原式$= \dfrac{4a}{2a - b} - \dfrac{2b}{2a - b}$，
由于两个分式的分母相同，可以直接进行分子的加减运算，即：
$= \dfrac{4a - 2b}{2a - b}$，
提取公因子2，得：
$= \dfrac{2(2a - b)}{2a - b}$，
由于分子和分母有相同的因子$2a - b$（且$2a - b \neq 0$），可以约去，得：
$= 2$。
由于两个分式的分母相同，可以直接进行分子的加减运算，即：
$= \dfrac{4a - 2b}{2a - b}$，
提取公因子2，得：
$= \dfrac{2(2a - b)}{2a - b}$，
由于分子和分母有相同的因子$2a - b$（且$2a - b \neq 0$），可以约去，得：
$= 2$。
2. 化简$\dfrac{a^{2}}{a - b}+\dfrac{b^{2}}{b - a}$的结果是(
A.$a + b$
B.$a - b$
C.$a^{2}-b^{2}$
D.1
A
)A.$a + b$
B.$a - b$
C.$a^{2}-b^{2}$
D.1
答案
A
解析
原式$=\dfrac{a^{2}}{a - b}-\dfrac{b^{2}}{a - b}=\dfrac{a^{2}-b^{2}}{a - b}=\dfrac{(a + b)(a - b)}{a - b}=a + b$
3. 化简：$\dfrac{1}{x - 1}-\dfrac{1}{x + 1}= $
$\dfrac{2}{x^2 - 1}$
.答案
$\dfrac{2}{x^2 - 1}$
解析
原式$=\dfrac{(x + 1) - (x - 1)}{(x - 1)(x + 1)}=\dfrac{x + 1 - x + 1}{x^2 - 1}=\dfrac{2}{x^2 - 1}$
4. 已知实数$a$,$b满足ab = 1$的两根,则$\dfrac{1}{a^{2}+1}+\dfrac{1}{b^{2}+1}= $
1
.答案
1
解析
因为$ab = 1$，所以$a^2b^2=(ab)^2=1$。
$\begin{aligned}\frac{1}{a^2 + 1} + \frac{1}{b^2 + 1}&=\frac{(b^2 + 1)+(a^2 + 1)}{(a^2 + 1)(b^2 + 1)}\\&=\frac{a^2 + b^2 + 2}{a^2b^2 + a^2 + b^2 + 1}\\&=\frac{a^2 + b^2 + 2}{1 + a^2 + b^2 + 1}\\&=\frac{a^2 + b^2 + 2}{a^2 + b^2 + 2}\\&=1\end{aligned}$
$\begin{aligned}\frac{1}{a^2 + 1} + \frac{1}{b^2 + 1}&=\frac{(b^2 + 1)+(a^2 + 1)}{(a^2 + 1)(b^2 + 1)}\\&=\frac{a^2 + b^2 + 2}{a^2b^2 + a^2 + b^2 + 1}\\&=\frac{a^2 + b^2 + 2}{1 + a^2 + b^2 + 1}\\&=\frac{a^2 + b^2 + 2}{a^2 + b^2 + 2}\\&=1\end{aligned}$
5. 计算：
(1)$\dfrac{5}{6a^{2}c}+\dfrac{1}{6a^{2}c}-\dfrac{2}{6ca^{2}}$；
(2)$\dfrac{x^{2}+9x}{x^{2}+3x}+\dfrac{x^{2}-9}{x^{2}+6x + 9}$；
(3)$\dfrac{a^{2}-5}{a - 2}+\dfrac{a}{2 - a}+\dfrac{1 + a}{a - 2}$；
(4)$\dfrac{a^{2}}{a + 1}-a + 1$.
(1)$\dfrac{5}{6a^{2}c}+\dfrac{1}{6a^{2}c}-\dfrac{2}{6ca^{2}}$；
(2)$\dfrac{x^{2}+9x}{x^{2}+3x}+\dfrac{x^{2}-9}{x^{2}+6x + 9}$；
(3)$\dfrac{a^{2}-5}{a - 2}+\dfrac{a}{2 - a}+\dfrac{1 + a}{a - 2}$；
(4)$\dfrac{a^{2}}{a + 1}-a + 1$.
答案
(1)
$\dfrac{5}{6a^{2}c}+\dfrac{1}{6a^{2}c}-\dfrac{2}{6ca^{2}}$
$=\dfrac{5 + 1 - 2}{6a^{2}c}$
$=\dfrac{4}{6a^{2}c}$
$=\dfrac{2}{3a^{2}c}$
(2)
$\dfrac{x^{2}+9x}{x^{2}+3x}+\dfrac{x^{2}-9}{x^{2}+6x + 9}$
$=\dfrac{x(x + 9)}{x(x + 3)}+\dfrac{(x + 3)(x - 3)}{(x + 3)^{2}}$
$=\dfrac{x + 9}{x + 3}+\dfrac{x - 3}{x + 3}$
$=\dfrac{x + 9+x - 3}{x + 3}$
$=\dfrac{2x + 6}{x + 3}$
$=\dfrac{2(x + 3)}{x + 3}$
$= 2$
(3)
$\dfrac{a^{2}-5}{a - 2}+\dfrac{a}{2 - a}+\dfrac{1 + a}{a - 2}$
$=\dfrac{a^{2}-5}{a - 2}-\dfrac{a}{a - 2}+\dfrac{1 + a}{a - 2}$
$=\dfrac{a^{2}-5 - a+1 + a}{a - 2}$
$=\dfrac{a^{2}-4}{a - 2}$
$=\dfrac{(a + 2)(a - 2)}{a - 2}$
$=a + 2$
(4)
$\dfrac{a^{2}}{a + 1}-a + 1$
$=\dfrac{a^{2}}{a + 1}-(a - 1)$
$=\dfrac{a^{2}}{a + 1}-\dfrac{(a - 1)(a + 1)}{a + 1}$
$=\dfrac{a^{2}-(a^{2}-1)}{a + 1}$
$=\dfrac{a^{2}-a^{2}+1}{a + 1}$
$=\dfrac{1}{a + 1}$
$\dfrac{5}{6a^{2}c}+\dfrac{1}{6a^{2}c}-\dfrac{2}{6ca^{2}}$
$=\dfrac{5 + 1 - 2}{6a^{2}c}$
$=\dfrac{4}{6a^{2}c}$
$=\dfrac{2}{3a^{2}c}$
(2)
$\dfrac{x^{2}+9x}{x^{2}+3x}+\dfrac{x^{2}-9}{x^{2}+6x + 9}$
$=\dfrac{x(x + 9)}{x(x + 3)}+\dfrac{(x + 3)(x - 3)}{(x + 3)^{2}}$
$=\dfrac{x + 9}{x + 3}+\dfrac{x - 3}{x + 3}$
$=\dfrac{x + 9+x - 3}{x + 3}$
$=\dfrac{2x + 6}{x + 3}$
$=\dfrac{2(x + 3)}{x + 3}$
$= 2$
(3)
$\dfrac{a^{2}-5}{a - 2}+\dfrac{a}{2 - a}+\dfrac{1 + a}{a - 2}$
$=\dfrac{a^{2}-5}{a - 2}-\dfrac{a}{a - 2}+\dfrac{1 + a}{a - 2}$
$=\dfrac{a^{2}-5 - a+1 + a}{a - 2}$
$=\dfrac{a^{2}-4}{a - 2}$
$=\dfrac{(a + 2)(a - 2)}{a - 2}$
$=a + 2$
(4)
$\dfrac{a^{2}}{a + 1}-a + 1$
$=\dfrac{a^{2}}{a + 1}-(a - 1)$
$=\dfrac{a^{2}}{a + 1}-\dfrac{(a - 1)(a + 1)}{a + 1}$
$=\dfrac{a^{2}-(a^{2}-1)}{a + 1}$
$=\dfrac{a^{2}-a^{2}+1}{a + 1}$
$=\dfrac{1}{a + 1}$
