1. 计算$(\frac {3y}{-2x})^{2}\cdot (\frac {2x}{3y})^{3}$的结果是 (
C
)
A.$\frac {x}{3y}$
B.$-\frac {x}{3y}$
C.$\frac {2x}{3y}$
D.$-\frac {2x}{3y}$
答案：C
解析：
$(\frac{3y}{-2x})^{2} \cdot (\frac{2x}{3y})^{3}$
$=\frac{(3y)^{2}}{(-2x)^{2}} \cdot \frac{(2x)^{3}}{(3y)^{3}}$
$=\frac{9y^{2}}{4x^{2}} \cdot \frac{8x^{3}}{27y^{3}}$
$=\frac{9y^{2} \cdot 8x^{3}}{4x^{2} \cdot 27y^{3}}$
$=\frac{72x^{3}y^{2}}{108x^{2}y^{3}}$
$=\frac{2x}{3y}$
C
2. 若$\frac {□}{x+y}÷\frac {x}{y^{2}-x^{2}}$的结果为整式,则“□”中的式子可能是 (
C
)
A.$y-x$
B.$y+x$
C.$2x$
D.$\frac {1}{x}$
答案：C
解析：
$\begin{aligned}&\frac{□}{x+y}÷\frac{x}{y^2 - x^2}\\=&\frac{□}{x+y}×\frac{(y - x)(y + x)}{x}\\=&\frac{□(y - x)}{x}\end{aligned}$
要使结果为整式，则$□(y - x)$能被$x$整除。
选项C：$□ = 2x$时，$\frac{2x(y - x)}{x}=2(y - x)$，为整式。
C
3. 计算:(1)$\frac {a}{b^{2}}\cdot (-\frac {b}{a^{2}})=$
$-\frac {1}{ab}$
; (2)$(xy-x^{2})\cdot \frac {xy}{x-y}=$
$-x^{2}y$
;
(3)$\frac {a^{2}-1}{a^{2}+2a}÷\frac {a-1}{a}=$
$\frac {a+1}{a+2}$
; (4)$\frac {1}{x-1}\cdot \frac {x^{2}-2x+1}{x+1}=$
$\frac {x-1}{x+1}$
.
答案：
(1)$-\frac {1}{ab}$
(2)$-x^{2}y$
(3)$\frac {a+1}{a+2}$
(4)$\frac {x-1}{x+1}$
4. (2024春·雁塔区月考)化简:$\frac {a}{a+1}÷\frac {a^{2}}{a^{2}-1}=$
$\frac {a-1}{a}$
.
答案：$\frac {a-1}{a}$
解析：
$\frac{a}{a+1} ÷ \frac{a^2}{a^2 - 1}$
$= \frac{a}{a+1} × \frac{(a+1)(a-1)}{a^2}$
$= \frac{a-1}{a}$
5. 计算:
(1)$(-\frac {a}{b^{2}c})\cdot \frac {bc^{2}}{a}$; (2)$\frac {x+2}{x-3}\cdot \frac {x^{2}-6x+9}{x^{2}-4}$; (3)$\frac {y^{2}}{6x}÷\frac {1}{3x^{2}}$; (4)$(a^{2}-a)÷\frac {a}{a-1}$.
答案：解:
(1)原式$=-\frac {abc^{2}}{a^{2}b^{2}c}=-\frac {c}{b}.$
(2)原式$=\frac {x+2}{x-3}\cdot \frac {(x-3)^{2}}{(x+2)(x-2)}=\frac {x-3}{x-2}.$
(3)原式$=\frac {y^{2}}{6x}\cdot 3x^{2}=\frac {xy^{2}}{2}.$
(4)原式$=a(a-1)\cdot \frac {a-1}{a}=(a-1)^{2}.$
6. 先化简,再求值:$\frac {4}{x^{2}-4}÷\frac {2}{x-2}$,其中$x= 1$.
答案：解:原式$=\frac {4}{(x+2)(x-2)}\cdot \frac {x-2}{2}=\frac {2}{x+2}.$当$x=1$时,原式$=\frac {2}{1+2}=\frac {2}{3}.$
7. 关于式子$\frac {x^{2}+2x+1}{x^{2}-1}÷\frac {x}{x-1}$,下列说法正确的是 (
D
)
A.当$x= 1$时,其值为2
B.当$x= -1$时,其值为0
C.当$-1\lt x\lt0$时,其值为正数
D.当$x\lt-1$时,其值为正数
答案：D
解析：
$\begin{aligned}&\frac{x^2 + 2x + 1}{x^2 - 1} ÷ \frac{x}{x - 1}\\=&\frac{(x + 1)^2}{(x + 1)(x - 1)} \cdot \frac{x - 1}{x}\\=&\frac{x + 1}{x} \quad (x \neq \pm 1, 0)\\\\&A. x = 1 \text{时，原式无意义，错误；}\\&B. x = -1 \text{时，原式无意义，错误；}\\&C. -1 ＜ x ＜ 0 \text{时，} x + 1 ＞ 0, x ＜ 0, \frac{x + 1}{x} ＜ 0 \text{，错误；}\\&D. x ＜ -1 \text{时，} x + 1 ＜ 0, x ＜ 0, \frac{x + 1}{x} ＞ 0 \text{，正确.}\end{aligned}$
D
8. 下列计算正确的是 (
D
)
A.$(-a)^{3}= a^{3}$
B.$2a^{2}+a^{3}= 3a^{5}$
C.$a^{6}÷a^{3}= a^{2}$
D.$(-\frac {b^{2}}{2a})^{3}= -\frac {b^{6}}{8a^{3}}$
答案：D
解析：
A.$(-a)^{3}=-a^{3}\neq a^{3}$
B.$2a^{2}$与$a^{3}$不是同类项，不能合并
C.$a^{6}÷a^{3}=a^{6-3}=a^{3}\neq a^{2}$
D.$(-\frac{b^{2}}{2a})^{3}=-\frac{(b^{2})^{3}}{(2a)^{3}}=-\frac{b^{6}}{8a^{3}}$
结论：D
9. 计算:(1)$\frac {2}{m^{2}-m}÷\frac {1}{m-1}=$
$\frac {2}{m}$
; (2)$\frac {x^{2}-2x+1}{x^{2}-1}÷\frac {x^{2}-x}{x+1}=$
$\frac {1}{x}$
.
答案：
(1)$\frac {2}{m}$
(2)$\frac {1}{x}$
