1[中]若$ab=1$，$m=\dfrac{1}{1+a}+\dfrac{1}{1+b}$，则$m^{2021}$的值为（）

A.2021
B.0
C.1
D.2

1. C 【解析】$\because ab = 1, \therefore m = \dfrac{1}{1+a} + \dfrac{1}{1+b} = \dfrac{1+b+1+a}{(1+a)(1+b)} = \dfrac{2+b+a}{1+a+b+ab} = \dfrac{2+b+a}{1+a+b+1} = 1,$
$\therefore m^{2021} = 1^{2021} = 1.$ 故选 C.

2[中]已知$x+y=5$，$xy=2$，则$\frac{x^2 + 3xy + y^2}{x^2y + xy^2}$的值为（）

A.2
B.$\frac{9}{4}$
C.3
D.$\frac{27}{10}$

2. D 【解析】原式$=\dfrac{(x+y)^2+xy}{xy(x+y)}$, 把$x+y=5,xy=2$代入得原式$=\dfrac{25+2}{2×5}=\dfrac{27}{10}.$ 故选 D.

3[中]若$y=\frac{x}{1-2x}$，则$\frac{2x-3xy-2y}{y+xy-x}$的值为（）

A.$\frac{1}{3}$
B.$-1$
C.$-\frac{5}{3}$
D.$-\frac{7}{3}$

3. D 【解析】$\because y=\dfrac{x}{1-2x},\therefore y-2xy=x,\therefore y-x=2xy,\therefore \dfrac{2x-3xy-2y}{y+xy-x}=\dfrac{2x-2y-3xy}{y-x+xy}=\dfrac{-2(y-x)-3xy}{y-x+xy}=\dfrac{-7xy}{3xy}=-\dfrac{7}{3},$ 故选 D.

4[2026 四川眉山期中,中]先化简,再求值:
$( \dfrac{a+2}{a-2} + \dfrac{1}{a-2} ) · (a^2 - 2a)$,其中 $a^2 + 3a - 5 = 0.$

4.【解】原式$=\dfrac{a+3}{a-2} · a(a-2) = a(a+3) = a^2+3a.$
由$a^2+3a-5=0$可得$a^2+3a=5,\therefore$ 原式$=5.$

5[中]已知$(x-y)(2x-y)=0(xy≠0)$，则$\frac{x}{y}+\frac{y}{x}$的值是（）

A.2
B.$-2\frac{1}{2}$
C.$-2$或$-2\frac{1}{2}$
D.2或$2\frac{1}{2}$

5. D 【解析】$\because (x-y)(2x-y)=0(xy≠0),\therefore x-y=0$或$2x-y=0$,解得$x=y$或$2x=y$. 原式$=\dfrac{(x+y)^2-2xy}{xy}=\dfrac{(x+y)^2}{xy}-2$. 当$x=y$时,原式$=\dfrac{4y^2}{y^2}-2=4-2=2$;当$2x=y$时,原式$=\dfrac{9x^2}{2x^2}-2=\dfrac{9}{2}-2=2\dfrac{1}{2},\therefore$ 原式的值是2或$2\dfrac{1}{2}$.

6[中]先化简,再求值:$(1-\dfrac{3a-10}{a-2})÷\dfrac{a-4}{a^2-4a+4}$,其中$|a|≤4$,且$a$为整数.

6.【解】原式 $= (\dfrac{a-2}{a-2}-\dfrac{3a-10}{a-2}) ÷ \dfrac{a-4}{(a-2)^2} = \dfrac{a-2-3a+10}{a-2} · \dfrac{(a-2)^2}{a-4} = \dfrac{-2(a-4)}{a-2} · \dfrac{(a-2)^2}{a-4} = -2a+4.$ $\because |a| ≤ 4$, 且$a$为整数,$\therefore a = \pm 4,\pm3,\pm2,\pm1,0.$ 由分式有意义的条件可知,$a$不能取2和4. 当$a=0$时,原式$=0+4=4$;当$a=1$时,原式$=-2+4=2$;当$a=-1$时,原式$=2+4=6$;当$a=-2$时,原式$=4+4=8$;当$a=3$时,原式$=-6+4=-2$;当$a=-3$时,原式$=6+4=10$;当$a=-4$时,原式$=8+4=12.$

7[2026江苏连云港校级期中,中]在初中数学学习阶段,我们常常会利用一些变形技巧来简化式子,从而解答问题.
材料:在解决某些分式问题时,倒数法是常用的变形技巧之一,所谓倒数法,即把式子变成其倒数形式,再通过约分进行化简,从而达到计算目的.
例:若$\frac{x}{x^2+1}=\frac{1}{4}$,求代数式$x^2+\frac{1}{x^2}$的值.
解:$\because \frac{x}{x^2+1}=\frac{1}{4},\therefore \frac{x^2+1}{x}=4$,即$\frac{x^2}{x}+\frac{1}{x}=4$,
$\therefore x+\frac{1}{x}=4,\therefore x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2 -2=16-2=14$.
根据材料回答问题:
(1)已知$\frac{x}{x^2 -x +1}=\frac{1}{4}$,求$x^2+\frac{1}{x^2}$的值;
(2)已知$x,y,z$为实数,$\frac{xy}{x+y}=-2,\frac{yz}{y+z}=\frac{4}{3},\frac{zx}{z+x}=-\frac{4}{3}$,求分式$\frac{xyz}{xy+yz+zx}$的值.

7.【解】(1)由条件可得$\dfrac{x^2-x+1}{x}=4$,即$\dfrac{x^2}{x}-\dfrac{x}{x}+\dfrac{1}{x}=4,\therefore \dfrac{x^2}{x}+\dfrac{1}{x}=5,\therefore x+\dfrac{1}{x}=5,\therefore x^2+\dfrac{1}{x^2}=(x+\dfrac{1}{x})^2-2=25-2=23.$
(2)由条件可得$\dfrac{x+y}{xy}=-\dfrac{1}{2},\therefore \dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{2}.$ 同理可得$\dfrac{1}{x}+\dfrac{1}{z}=-\dfrac{3}{4},\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{4},\therefore \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}=-\dfrac{1}{2}+(-\dfrac{3}{4})+\dfrac{3}{4}=-\dfrac{1}{2},\therefore \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-\dfrac{1}{4},\therefore \dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-\dfrac{1}{4},\therefore \dfrac{xyz}{xy+yz+zx}=-4.$