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1. 下列计算正确的是（
A.$(-\dfrac{b}{2a^2})^3 = -\dfrac{b^3}{6a^6}$
B.$\dfrac{-x}{x-y} = \dfrac{x}{-x+y}$
C.$(-\dfrac{x^2}{y})^2 ÷ (-\dfrac{y}{x})^3 = -xy$
D.$\dfrac{0.2a + b}{0.7a - b} = \dfrac{2a + b}{7a - b}$
B
）A.$(-\dfrac{b}{2a^2})^3 = -\dfrac{b^3}{6a^6}$
B.$\dfrac{-x}{x-y} = \dfrac{x}{-x+y}$
C.$(-\dfrac{x^2}{y})^2 ÷ (-\dfrac{y}{x})^3 = -xy$
D.$\dfrac{0.2a + b}{0.7a - b} = \dfrac{2a + b}{7a - b}$
答案
B
2. 计算$(\dfrac{a-b}{b})^2 · \dfrac{b}{a^2 - b^2}$的结果是 （
A.$\dfrac{1}{b}$
B.$\dfrac{a-b}{ab + b^2}$
C.$\dfrac{a-b}{a + b}$
D.$\dfrac{1}{b(a + b)}$
B
）A.$\dfrac{1}{b}$
B.$\dfrac{a-b}{ab + b^2}$
C.$\dfrac{a-b}{a + b}$
D.$\dfrac{1}{b(a + b)}$
答案
B
3. [母题 教材 P149 例5] 计算$\frac{x^2}{y} ÷ (-\frac{y}{x}) · (\frac{y}{x})^2$的结果是（
A.$-x$
B.$-\frac{x^2}{y}$
C.$\frac{x}{y}$
D.$\frac{x^2}{y}$
A
）A.$-x$
B.$-\frac{x^2}{y}$
C.$\frac{x}{y}$
D.$\frac{x^2}{y}$
答案
A
4. 化简$\frac{x^2 -1}{x^2 -2x +1} ÷ \frac{x+1}{x-1} · \frac{1-x}{1+x}$的结果为（
A.$\frac{x+1}{x-1}$
B.$\frac{x-1}{x+1}$
C.$\frac{1-x}{1+x}$
D.$\frac{1+x}{1-x}$
C
）A.$\frac{x+1}{x-1}$
B.$\frac{x-1}{x+1}$
C.$\frac{1-x}{1+x}$
D.$\frac{1+x}{1-x}$
答案
C
5. 计算$[(-\dfrac{n}{ab})^2]^3$与$-(\dfrac{n^3}{a^3b^3})^2$的结果可知，它们（
A.相等
B.互为倒数
C.互为相反数
D.以上都不对
C
）A.相等
B.互为倒数
C.互为相反数
D.以上都不对
答案
C
6.若$□ × ( \dfrac{y^2}{x} )^2 = \dfrac{y}{x}$，则$□$中的式子是
$\dfrac{x}{y^3}$
。答案
$\dfrac{x}{y^3}$
7. 已知$(\dfrac{x^3}{y^2})^2 ÷ (-\dfrac{x}{y^3})^2 =6$，则$x^4y^2$的值为
6
。答案
6
【点拨】$\because (\dfrac{x^3}{y^2})^2 ÷ (-\dfrac{x}{y^3})^2 = 6, \therefore \dfrac{x^6}{y^4} ÷ \dfrac{x^2}{y^6} = 6, \therefore \dfrac{x^6}{y^4} · \dfrac{y^6}{x^2}=6, \therefore x^4 y^2=6.$
【点拨】$\because (\dfrac{x^3}{y^2})^2 ÷ (-\dfrac{x}{y^3})^2 = 6, \therefore \dfrac{x^6}{y^4} ÷ \dfrac{x^2}{y^6} = 6, \therefore \dfrac{x^6}{y^4} · \dfrac{y^6}{x^2}=6, \therefore x^4 y^2=6.$
8. 母题 教材 P151 习题 T4 计算：
(1)[2025东营月考] $(-\dfrac{2a}{b^2})^2 · (\dfrac{2b}{a})^2 ÷ (-\dfrac{2b^2}{a})$；
(2) $(\dfrac{x^2 - y^2}{xy})^2 ÷ (x+y) · (\dfrac{x}{x-y})^3$。
(1)[2025东营月考] $(-\dfrac{2a}{b^2})^2 · (\dfrac{2b}{a})^2 ÷ (-\dfrac{2b^2}{a})$；
(2) $(\dfrac{x^2 - y^2}{xy})^2 ÷ (x+y) · (\dfrac{x}{x-y})^3$。
答案
(1)$(-\dfrac{2a}{b^2})^2 · (\dfrac{2b}{a})^2 ÷ (-\dfrac{2b^2}{a})= -\dfrac{4a^2}{b^4} · \dfrac{4b^2}{a^2} · \dfrac{a}{2b^2}= -\dfrac{8a}{b^4}.$
(2)原式$=\dfrac{(x+y)^2(x-y)^2}{x^2 y^2} × \dfrac{1}{x+y} × \dfrac{x^3}{(x-y)^3} = \dfrac{x(x+y)}{y^2 (x-y)} = \dfrac{x^2 + xy}{x y^2 - y^3}.$
(2)原式$=\dfrac{(x+y)^2(x-y)^2}{x^2 y^2} × \dfrac{1}{x+y} × \dfrac{x^3}{(x-y)^3} = \dfrac{x(x+y)}{y^2 (x-y)} = \dfrac{x^2 + xy}{x y^2 - y^3}.$
9. 已知$a^2 + b^2 - 10a - 6b + 34 = 0$，求$(\dfrac{a - b}{a + b})^2 ÷ \dfrac{a^2 - b^2}{ab} · \dfrac{a^2 + 2ab + b^2}{a^2 - ab}$的值。
答案
$(\dfrac{a-b}{a+b})^2 ÷ \dfrac{a^2 - b^2}{ab} · \dfrac{a^2 + 2ab + b^2}{a^2 - ab} = \dfrac{(a-b)^2}{(a+b)^2} · \dfrac{ab}{(a+b)(a-b)} · \dfrac{(a+b)^2}{a(a-b)} = \dfrac{b}{a+b}.$
$\because a^2 + b^2 -10a -6b +34 = 0,$
$\therefore (a-5)^2 + (b-3)^2 = 0.$
$\because (a-5)^2 \ge 0, (b-3)^2 \ge 0,$
$\therefore a-5=0,b-3=0,$
$\therefore a=5,b=3,$
$\therefore 原式=\dfrac{b}{a+b}=\dfrac{3}{5+3}=\dfrac{3}{8}.$
$\because a^2 + b^2 -10a -6b +34 = 0,$
$\therefore (a-5)^2 + (b-3)^2 = 0.$
$\because (a-5)^2 \ge 0, (b-3)^2 \ge 0,$
$\therefore a-5=0,b-3=0,$
$\therefore a=5,b=3,$
$\therefore 原式=\dfrac{b}{a+b}=\dfrac{3}{5+3}=\dfrac{3}{8}.$
10. 彤彤做错了下列计算题中的一道题，你认为她做错的题是（
A.$\frac{y}{x^2} ÷ \frac{(-y)^2}{x} = \frac{1}{xy}$
B.$( \frac{3x}{-2y} )^2 · ( \frac{2y}{-3x} )^3 = -\frac{2y}{3x}$
C.$\frac{xy}{x^2 - 2xy + y^2} ÷ \frac{xy^2 + x^2 y}{x^2 - y^2} = \frac{1}{x - y}$
D.$\frac{x^2 + x}{x^2 + 2x + 1} · \frac{x^2 - 1}{x - 1} = x(x + 1)$
D
）A.$\frac{y}{x^2} ÷ \frac{(-y)^2}{x} = \frac{1}{xy}$
B.$( \frac{3x}{-2y} )^2 · ( \frac{2y}{-3x} )^3 = -\frac{2y}{3x}$
C.$\frac{xy}{x^2 - 2xy + y^2} ÷ \frac{xy^2 + x^2 y}{x^2 - y^2} = \frac{1}{x - y}$
D.$\frac{x^2 + x}{x^2 + 2x + 1} · \frac{x^2 - 1}{x - 1} = x(x + 1)$
答案
D
