1.(2025·西安期末)若$(1-\dfrac{4}{x+1})□\dfrac{x-3}{x^2-1}$的结果是$x-1$，则$□$处的运算符号是（
D
）
A.＋
B.－
C.×
D.÷
答案：1.D
2. 计算：
(1)$\frac{x^2 -1}{x+2} ÷ (x+1)$；
(2)
；
(3)$1 - \frac{a-2}{a} ÷ \frac{a^2 -4}{a^2 +a}$；
(4)
；
答案：2.解:(1)原式$=\frac{(x+1)(x-1)}{x+2} · \frac{1}{x+1} =\frac{x-1}{x+2}$.
(2)原式$=\frac{(m+3)(m-3)}{(m+3)^2} ÷ \frac{m+3-m}{m+3} =\frac{m-3}{m+3} · \frac{m+3}{3} =\frac{m-3}{3}$.
(3)原式$=1-\frac{a-2}{a} ÷ \frac{(a+2)(a-2)}{a(a+1)} =1-\frac{a-2}{a} · \frac{a(a+1)}{(a+2)(a-2)} =1-\frac{a+1}{a+2} =\frac{a+2-(a+1)}{a+2} =\frac{1}{a+2}$.
(4)原式$=\frac{m}{(m-1)(m+1)} ÷ \frac{m(m-1)}{(m-1)^2} =\frac{m}{(m-1)(m+1)} · \frac{(m-1)^2}{m(m-1)} =\frac{1}{m+1}$.
3. 先化简，再求值：
(1)(2025·苏州)$(\dfrac{2}{x-1}+1)·\dfrac{x^2 - x}{x^2 + 2x + 1}$，其中$x=-2$；
(2)(2025·眉山)$(\dfrac{y}{x^2 - y^2}+\dfrac{1}{x + y})÷\dfrac{x}{x - y}$，其中$x,y$满足$(x+2)^2 + |y - 1|=0$。
答案：3.解:(1)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$=\frac{-2}{-2+1}=2$.
(2)原式$=[\frac{y}{(x+y)(x-y)}+\frac{x-y}{(x+y)(x-y)}] · \frac{x-y}{x} =\frac{x}{(x+y)(x-y)} · \frac{x-y}{x} =\frac{1}{x+y}$.
$\because (x+2)^2+|y-1|=0$,
$\therefore x+2=0,y-1=0$,解得$x=-2,y=1$,
$\therefore$原式$=\frac{1}{-2+1}=-1$.
4.(2025·滕州月考)化简$\frac{a^2 -4}{a^2 +2a +1} ÷ \frac{a^2 -4a +4}{(a+1)^2} - \frac{2}{a-2}$的结果为（
C
）
A.$\frac{a+2}{a-2}$
B.$\frac{a-4}{a-2}$
C.$\frac{a}{a-2}$
D.$a$
答案：4.C
5.(2024·北京模拟)若$3ab - 3b^2 - 2 = 0$，则代数式$(1 - \dfrac{2ab - b^2}{a^2}) ÷ \dfrac{a - b}{a^2b}$的值为
$\frac{2}{3}$
.
答案：5.$\frac{2}{3}$
6. 计算:
(1) $(a - 1 + \dfrac{1}{a - 3}) ÷ \dfrac{a^2 - 4}{a - 3}$;
(2) $(1 + \dfrac{1}{x}) ÷ (2x - \dfrac{1 + x^2}{x})$;
(3) $(\dfrac{a}{a - 2} - \dfrac{4}{a^2 - 2a}) ÷ \dfrac{a + 2}{a}$;
(4) $\dfrac{x^2 + 2x + 1}{2x - 6} ÷ (1 + \dfrac{4}{x - 3})$.
答案：6.解:(1)原式$=[\frac{(a-1)(a-3)}{a-3}+\frac{1}{a-3}] ÷ \frac{(a+2)(a-2)}{a-3}$
$=(\frac{a^2-4a+3}{a-3}+\frac{1}{a-3}) · \frac{a-3}{(a+2)(a-2)} =\frac{(a-2)^2}{a-3} · \frac{a-3}{(a+2)(a-2)} =\frac{a-2}{a+2}$.
(2)原式$=\frac{x+1}{x} ÷ \frac{2x^2-1-x^2}{x} =\frac{x+1}{x} · \frac{x}{(x+1)(x-1)} =\frac{1}{x-1}$.
(3)原式$=[\frac{a}{a-2}-\frac{4}{a(a-2)}] · \frac{a}{a+2} =\frac{(a+2)(a-2)}{a(a-2)} · \frac{a}{a+2} =1$.
(4)原式$=\frac{(x+1)^2}{2(x-3)} ÷ \frac{x-3+4}{x-3} =\frac{(x+1)^2}{2(x-3)} · \frac{x-3}{x+1} =\frac{x+1}{2}$.
