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1.「2025江苏苏州中考」先化简，再求值：$(\dfrac{2}{x-1}+1) · \dfrac{x^2 - x}{x^2 + 2x + 1}$，其中$x=-2$。
答案
1. 解析 原式=$\frac{2+x-1}{x-1} · \frac{x(x-1)}{(x+1)^2} = \frac{x+1}{x-1} · \frac{x(x-1)}{(x+1)^2} = \frac{x}{x+1}$.
当$x=-2$时,原式=$\frac{-2}{-2+1}=2$.
当$x=-2$时,原式=$\frac{-2}{-2+1}=2$.
2.「2026甘肃嘉峪关期末」先化简,再求值:$\frac{2}{m+1} - \frac{m-2}{m^2 -1} ÷ (1 - \frac{1}{m^2 -2m +1})$,其中$m=3$.
答案
2. 解析 原式=$\frac{2}{m+1} - \frac{m-2}{m^2-1} ÷ \frac{m^2-2m+1-1}{m^2-2m+1}$
=$\frac{2}{m+1} - \frac{m-2}{(m-1)(m+1)} · \frac{(m-1)^2}{m(m-2)}$
=$\frac{2}{m+1} - \frac{m-1}{m(m+1)}$
=$\frac{2m -m +1}{m(m+1)} = \frac{m+1}{m(m+1)} = \frac{1}{m}$.
当$m=3$时,原式=$\frac{1}{3}$.
=$\frac{2}{m+1} - \frac{m-2}{(m-1)(m+1)} · \frac{(m-1)^2}{m(m-2)}$
=$\frac{2}{m+1} - \frac{m-1}{m(m+1)}$
=$\frac{2m -m +1}{m(m+1)} = \frac{m+1}{m(m+1)} = \frac{1}{m}$.
当$m=3$时,原式=$\frac{1}{3}$.
3.「2026上海松江期末改编」已知$(a+2)^2 + |b-1|=0$，则$( \frac{a^2 - b^2}{a^2 - 2ab + b^2} + \frac{a}{b - a} ) ÷ \frac{b^2}{a^2 - ab}$的值为
-2
。答案
3. 答案 -2
解析 原式=$[\frac{(a+b)(a-b)}{(a-b)^2}+\frac{a}{b-a}] · \frac{a(a-b)}{b^2}$
=$(\frac{a+b}{a-b}-\frac{a}{a-b}) · \frac{a(a-b)}{b^2}$
=$\frac{b}{a-b} · \frac{a(a-b)}{b^2} = \frac{a}{b}$,
$\because (a+2)^2 + |b-1|=0,\therefore a=-2,b=1$,
当$a=-2,b=1$时,原式=$\frac{-2}{1}=-2$.
解析 原式=$[\frac{(a+b)(a-b)}{(a-b)^2}+\frac{a}{b-a}] · \frac{a(a-b)}{b^2}$
=$(\frac{a+b}{a-b}-\frac{a}{a-b}) · \frac{a(a-b)}{b^2}$
=$\frac{b}{a-b} · \frac{a(a-b)}{b^2} = \frac{a}{b}$,
$\because (a+2)^2 + |b-1|=0,\therefore a=-2,b=1$,
当$a=-2,b=1$时,原式=$\frac{-2}{1}=-2$.
4.「2026广东深圳模拟」先化简,再求值:$(\dfrac{x}{x^2 + x} -1) ÷ \dfrac{x^2 -1}{x^2 +2x +1}$, 其中$x$的值是不等式组$\begin{cases}3x+2＜-4, \\2(x-1)≤3x+1\end{cases}$的整数解.
答案
4. 解析 原式=$\frac{x-x^2-x}{x^2+x} · \frac{(x+1)^2}{(x+1)(x-1)}$
=$\frac{-x^2}{x(x+1)} · \frac{x+1}{x-1} = -\frac{x}{x-1}$,
解不等式组$\begin{cases}3x+2＜-4①, \\2(x-1)≤3x+1②,\end{cases}$
解不等式①得$x＜-2$,解不等式②得$x≥-3$,
∴原不等式组的解集为$-3≤x＜-2$,
∵$x$为整数,
∴$x$的值为$-3$,
当$x=-3$时,原式=$-\frac{-3}{-3-1}=-\frac{3}{4}$.
=$\frac{-x^2}{x(x+1)} · \frac{x+1}{x-1} = -\frac{x}{x-1}$,
解不等式组$\begin{cases}3x+2＜-4①, \\2(x-1)≤3x+1②,\end{cases}$
解不等式①得$x＜-2$,解不等式②得$x≥-3$,
∴原不等式组的解集为$-3≤x＜-2$,
∵$x$为整数,
∴$x$的值为$-3$,
当$x=-3$时,原式=$-\frac{-3}{-3-1}=-\frac{3}{4}$.
5.「2024四川广安中考」先化简$(a+1-\dfrac{3}{a-1})÷\dfrac{a^2+4a+4}{a-1}$，再从$-2,0,1,2$中选取一个适合的数代入求值.
答案
5. 解析 原式=$(\frac{a^2-1}{a-1}-\frac{3}{a-1}) · \frac{a-1}{a^2+4a+4}$
=$\frac{(a+2)(a-2)}{a-1} · \frac{a-1}{(a+2)^2} = \frac{a-2}{a+2}$,
由题意得$a≠1$且$a≠-2$,
当$a=0$时,原式=$\frac{0-2}{0+2}=-1$,当$a=2$时,原式=$\frac{2-2}{2+2}=0$.
=$\frac{(a+2)(a-2)}{a-1} · \frac{a-1}{(a+2)^2} = \frac{a-2}{a+2}$,
由题意得$a≠1$且$a≠-2$,
当$a=0$时,原式=$\frac{0-2}{0+2}=-1$,当$a=2$时,原式=$\frac{2-2}{2+2}=0$.
6.先化简,再求值:$\frac{y^2}{xy+2y^2}-\frac{1}{y-1}÷\frac{x+2y}{y^2-2y+1}$,其中$3x+6y-1=0.$
答案
6. 解析 原式=$\frac{y^2}{y(x+2y)} - \frac{1}{y-1} · \frac{(y-1)^2}{x+2y}$
=$\frac{y}{x+2y} - \frac{y-1}{x+2y} = \frac{1}{x+2y}$,
由$3x+6y-1=0$,得$x+2y=\frac{1}{3}$,
∴原式=$\frac{1}{x+2y} = \frac{1}{\frac{1}{3}}=3$.
=$\frac{y}{x+2y} - \frac{y-1}{x+2y} = \frac{1}{x+2y}$,
由$3x+6y-1=0$,得$x+2y=\frac{1}{3}$,
∴原式=$\frac{1}{x+2y} = \frac{1}{\frac{1}{3}}=3$.
7.「2026 山东潍坊月考」先化简,再求值:$\frac{x^2 -8x +16}{x^2 +2x} ÷ ( \frac{12}{x+2} -x +2 ) + \frac{1}{x+4}$,其中$x$满足$x^2 +4x -4=0$.
答案
7. 解析 原式=$\frac{(x-4)^2}{x(x+2)} ÷ [\frac{12}{x+2}-(x-2)] + \frac{1}{x+4}$
=$\frac{(x-4)^2}{x(x+2)} ÷ (\frac{12}{x+2} - \frac{x^2-4}{x+2}) + \frac{1}{x+4}$
=$\frac{(x-4)^2}{x(x+2)} ÷ \frac{16-x^2}{x+2} + \frac{1}{x+4}$
=$\frac{(4-x)^2}{x(x+2)} × \frac{x+2}{(4+x)(4-x)} + \frac{1}{x+4}$
=$\frac{4-x}{x(x+4)} + \frac{x}{x(x+4)} = \frac{4}{x(x+4)} = \frac{4}{x^2+4x}$,
$\because x^2+4x-4=0,\therefore x^2+4x=4,\therefore$原式=$\frac{4}{4}=1$.
=$\frac{(x-4)^2}{x(x+2)} ÷ (\frac{12}{x+2} - \frac{x^2-4}{x+2}) + \frac{1}{x+4}$
=$\frac{(x-4)^2}{x(x+2)} ÷ \frac{16-x^2}{x+2} + \frac{1}{x+4}$
=$\frac{(4-x)^2}{x(x+2)} × \frac{x+2}{(4+x)(4-x)} + \frac{1}{x+4}$
=$\frac{4-x}{x(x+4)} + \frac{x}{x(x+4)} = \frac{4}{x(x+4)} = \frac{4}{x^2+4x}$,
$\because x^2+4x-4=0,\therefore x^2+4x=4,\therefore$原式=$\frac{4}{4}=1$.
