1. 化简$\frac {m^{2}-n^{2}}{m^{2}}÷\frac {m-n}{m}$的结果是 (
A.$\frac {m}{m-n}$
B.$\frac {m}{m+n}$
C.$\frac {m-n}{m}$
D.$\frac {m+n}{m}$
D
)A.$\frac {m}{m-n}$
B.$\frac {m}{m+n}$
C.$\frac {m-n}{m}$
D.$\frac {m+n}{m}$
答案：D
解析：
$\begin{aligned}\frac{m^2 - n^2}{m^2} ÷ \frac{m - n}{m} &= \frac{(m + n)(m - n)}{m^2} × \frac{m}{m - n} \\&= \frac{(m + n)\cancel{(m - n)}}{m^2} × \frac{m}{\cancel{m - n}} \\&= \frac{m + n}{m}\end{aligned}$
D
D
2. 计算$(a^{2}b)^{3}÷\frac {a}{b^{2}}$的结果是 (
A.$a^{5}b^{5}$
B.$a^{4}b^{5}$
C.$ab^{5}$
D.$a^{5}b^{6}$
A
)A.$a^{5}b^{5}$
B.$a^{4}b^{5}$
C.$ab^{5}$
D.$a^{5}b^{6}$
答案：A
解析：
$(a^{2}b)^{3}÷\frac{a}{b^{2}}$
$=a^{6}b^{3}×\frac{b^{2}}{a}$
$=a^{5}b^{5}$
A
$=a^{6}b^{3}×\frac{b^{2}}{a}$
$=a^{5}b^{5}$
A
3. (2024·青山区三模)计算$\frac {x^{2}-2xy+y^{2}}{x^{2}}÷\frac {x-y}{x}$的结果为
$\frac{x-y}{x}$
。答案：$\frac{x-y}{x}$
解析：
$\frac{x^{2}-2xy+y^{2}}{x^{2}}÷\frac{x-y}{x}$
$=\frac{(x-y)^{2}}{x^{2}}×\frac{x}{x-y}$
$=\frac{x-y}{x}$
$\frac{x-y}{x}$
$=\frac{(x-y)^{2}}{x^{2}}×\frac{x}{x-y}$
$=\frac{x-y}{x}$
$\frac{x-y}{x}$
4. 当$x= 2,y= -3$时,式子$\frac {x^{2}-y^{2}}{x}\cdot \frac {x}{x^{2}+2xy+y^{2}}$的值为
-5
。答案：-5
解析：
$\begin{aligned}&\frac{x^2 - y^2}{x} \cdot \frac{x}{x^2 + 2xy + y^2}\\=&\frac{(x + y)(x - y)}{x} \cdot \frac{x}{(x + y)^2}\\=&\frac{x - y}{x + y}\\\end{aligned}$
当$x = 2$，$y = -3$时，
$\frac{2 - (-3)}{2 + (-3)} = \frac{5}{-1} = -5$
$-5$
当$x = 2$，$y = -3$时，
$\frac{2 - (-3)}{2 + (-3)} = \frac{5}{-1} = -5$
$-5$
5. 计算:
(1)$(-\frac {2b^{2}}{a^{3}})^{3};$
(2)$(-\frac {y}{x^{2}})^{3}\cdot (-\frac {x}{2y})^{2};$
(3)$(\frac {c^{3}}{a^{2}b})^{2}÷(\frac {c^{4}}{a^{3}b})^{2}÷(\frac {a}{c})^{4};$
(4)$(\frac {a-b}{ab})^{2}\cdot (\frac {-a}{b-a})^{3}\cdot (a^{2}-b^{2}).$
(1)$(-\frac {2b^{2}}{a^{3}})^{3};$
(2)$(-\frac {y}{x^{2}})^{3}\cdot (-\frac {x}{2y})^{2};$
(3)$(\frac {c^{3}}{a^{2}b})^{2}÷(\frac {c^{4}}{a^{3}b})^{2}÷(\frac {a}{c})^{4};$
(4)$(\frac {a-b}{ab})^{2}\cdot (\frac {-a}{b-a})^{3}\cdot (a^{2}-b^{2}).$
答案：5.解:
(1)原式$=-\frac{8b^{6}}{a^{9}}$.
(2)原式$=(-\frac{y^{3}}{x^{6}})\cdot \frac{x^{2}}{4y^{2}}=-\frac{y}{4x^{4}}$.
(3)原式$=\frac{c^{6}}{a^{4}b^{2}}\cdot \frac{a^{6}b^{2}}{c^{8}}\cdot \frac{c^{4}}{a^{4}}=\frac{c^{2}}{a^{2}}$.
(4)原式$=\frac{(a-b)^{2}}{a^{2}b^{2}}\cdot \frac{-a^{3}}{-(a-b)^{3}}\cdot (a-b)(a+b)=$ $\frac{a(a+b)}{b^{2}}$.
(1)原式$=-\frac{8b^{6}}{a^{9}}$.
(2)原式$=(-\frac{y^{3}}{x^{6}})\cdot \frac{x^{2}}{4y^{2}}=-\frac{y}{4x^{4}}$.
(3)原式$=\frac{c^{6}}{a^{4}b^{2}}\cdot \frac{a^{6}b^{2}}{c^{8}}\cdot \frac{c^{4}}{a^{4}}=\frac{c^{2}}{a^{2}}$.
(4)原式$=\frac{(a-b)^{2}}{a^{2}b^{2}}\cdot \frac{-a^{3}}{-(a-b)^{3}}\cdot (a-b)(a+b)=$ $\frac{a(a+b)}{b^{2}}$.
6. (2024春·淮安期末)若代数式$\frac {x-2}{x^{2}-4x+4}÷\frac {1}{x+6}$的值为F,则F的整数值有 (
A.0个
B.7个
C.8个
D.无数个
B
)A.0个
B.7个
C.8个
D.无数个
答案：B
解析：
$\begin{aligned}&\frac{x-2}{x^2 - 4x + 4} ÷ \frac{1}{x + 6}\\=&\frac{x - 2}{(x - 2)^2} \cdot (x + 6)\\=&\frac{x + 6}{x - 2} \quad (x \neq 2, x \neq -6)\\=&\frac{(x - 2) + 8}{x - 2} = 1 + \frac{8}{x - 2}\end{aligned}$
$\frac{8}{x - 2}$为整数，$x - 2$是8的因数：$\pm1,\pm2,\pm4,\pm8$
$x - 2 = 1$时，$F=9$；$x - 2=-1$时，$F=-7$；
$x - 2=2$时，$F=5$；$x - 2=-2$时，$F=-3$；
$x - 2=4$时，$F=3$；$x - 2=-4$时，$F=-1$；
$x - 2=8$时，$F=2$；$x - 2=-8$时，$F=0$；
共7个整数值，选B。
$\frac{8}{x - 2}$为整数，$x - 2$是8的因数：$\pm1,\pm2,\pm4,\pm8$
$x - 2 = 1$时，$F=9$；$x - 2=-1$时，$F=-7$；
$x - 2=2$时，$F=5$；$x - 2=-2$时，$F=-3$；
$x - 2=4$时，$F=3$；$x - 2=-4$时，$F=-1$；
$x - 2=8$时，$F=2$；$x - 2=-8$时，$F=0$；
共7个整数值，选B。
7. (2024春·朝阳区期中)化简$\frac {a^{2}-9}{a}\cdot \frac {1}{a-3}$的结果为
$\frac{a+3}{a}$
。答案：$\frac{a+3}{a}$
解析：
$\frac{a^2 - 9}{a} \cdot \frac{1}{a - 3} = \frac{(a + 3)(a - 3)}{a} \cdot \frac{1}{a - 3} = \frac{a + 3}{a}$
8. 计算:
(1)$\frac {a^{2}-16}{a^{2}+2a-8}÷(a-2)\cdot \frac {a^{2}+4-4a}{a-2};$
(2)$(xy-x^{2})^{2}÷\frac {(x-y)^{2}}{xy}\cdot \frac {x-y}{x^{2}};$
(3)$\frac {6-5x+x^{2}}{x^{2}-16}÷\frac {x-3}{4-x}\cdot \frac {x^{2}+5x+4}{4-x^{2}};$
(4)$(-\frac {2x}{y^{2}})^{3}\cdot (\frac {2y}{x})^{2}÷(-\frac {2y}{x}).$
(1)$\frac {a^{2}-16}{a^{2}+2a-8}÷(a-2)\cdot \frac {a^{2}+4-4a}{a-2};$
(2)$(xy-x^{2})^{2}÷\frac {(x-y)^{2}}{xy}\cdot \frac {x-y}{x^{2}};$
(3)$\frac {6-5x+x^{2}}{x^{2}-16}÷\frac {x-3}{4-x}\cdot \frac {x^{2}+5x+4}{4-x^{2}};$
(4)$(-\frac {2x}{y^{2}})^{3}\cdot (\frac {2y}{x})^{2}÷(-\frac {2y}{x}).$
答案：8.解:
(1)原式$=\frac{(a+4)(a-4)}{(a+4)(a-2)}\cdot \frac{1}{a-2}\cdot \frac{(a-2)^{2}}{a-2}=\frac{a-4}{a-2}$.
(2)原式$=x^{2}(x-y)^{2}\cdot \frac{xy}{(x-y)^{2}}\cdot \frac{x-y}{x^{2}}$ $=xy(x-y)=x^{2}y-xy^{2}$.
(3)原式$=\frac{(x-2)(x-3)}{(x+4)(x-4)}\cdot \frac{4-x}{x-3}\cdot \frac{(x+1)(x+4)}{(2+x)(2-x)}$ $=\frac{x+1}{x+2}$.
(4)原式$=-\frac{8x^{3}}{y^{5}}\cdot \frac{4y^{2}}{x^{2}}\cdot (-\frac{x}{2y})=\frac{16x^{2}}{y^{5}}$.
(1)原式$=\frac{(a+4)(a-4)}{(a+4)(a-2)}\cdot \frac{1}{a-2}\cdot \frac{(a-2)^{2}}{a-2}=\frac{a-4}{a-2}$.
(2)原式$=x^{2}(x-y)^{2}\cdot \frac{xy}{(x-y)^{2}}\cdot \frac{x-y}{x^{2}}$ $=xy(x-y)=x^{2}y-xy^{2}$.
(3)原式$=\frac{(x-2)(x-3)}{(x+4)(x-4)}\cdot \frac{4-x}{x-3}\cdot \frac{(x+1)(x+4)}{(2+x)(2-x)}$ $=\frac{x+1}{x+2}$.
(4)原式$=-\frac{8x^{3}}{y^{5}}\cdot \frac{4y^{2}}{x^{2}}\cdot (-\frac{x}{2y})=\frac{16x^{2}}{y^{5}}$.