《2025年一遍过八年级数学上册人教版》

1 [2025青岛莱西期中]计算$\frac {y}{x}÷\frac {y}{2}\cdot \frac {2}{y}$的结果是( )

A.$\frac {4}{xy}$
B.$\frac {1}{2}x$
C.$\frac {y}{x}$
D.$2y$

答案： A $\frac{y}{x}÷\frac{y}{2}\cdot\frac{2}{y}=\frac{y}{x}\cdot\frac{2}{y}\cdot\frac{2}{y}=\frac{4}{xy}$.

2 计算:$\frac {a^{2}-3a}{a^{2}+a}÷\frac {a-3}{a^{2}-1}\cdot \frac {a+1}{a-1}= $____.

答案： $a+1$ $\frac{a^2-3a}{a^2+a}÷\frac{a-3}{a^2-1}\cdot\frac{a+1}{a-1}=\frac{a(a-3)}{a(a+1)}÷\frac{a-3}{(a+1)(a-1)}\cdot$ $\frac{a+1}{a-1}=\frac{a(a-3)}{a(a+1)}\cdot\frac{(a+1)(a-1)}{a-3}\cdot\frac{a+1}{a-1}=a+1$.

3 [2025南通期中]先化简,再求值:$\frac {a+2}{a^{2}-2a+1}\cdot \frac {a^{2}-4a+4}{a+1}÷\frac {a^{2}-4}{a^{2}-1}$,其中$a= 3$.

答案： 解:$\frac{a+2}{a^2-2a+1}\cdot\frac{a^2-4a+4}{a+1}÷\frac{a^2-4}{a^2-1}$ $=\frac{a+2}{(a-1)^2}\cdot\frac{(a-2)^2}{a+1}\cdot\frac{(a+1)(a-1)}{(a+2)(a-2)}$ $=\frac{a-2}{a-1}$. 当$a=3$时,原式$=\frac{3-2}{3-1}=\frac{1}{2}$.

4 计算分式$(\frac {3y}{2x})^{2}$得( )

A.$\frac {3y}{2x}$
B.$\frac {3y}{2x^{2}}$
C.$\frac {9y^{2}}{4x^{2}}$
D.$\frac {9y^{2}}{2x}$

答案： C $(\frac{3y}{2x})^2=\frac{(3y)^2}{(2x)^2}=\frac{9y^2}{4x^2}$.

5 与$(-\frac {m+n}{3a^{2}})^{2}$相等的式子是( )

A.$-\frac {(m+n)^{2}}{6a^{2}}$
B.$\frac {(m+n)^{2}}{6a^{2}}$
C.$\frac {(m+n)^{2}}{9a^{4}}$
D.$\frac {m^{2}+n^{2}}{9a^{4}}$

答案： C $(-\frac{m+n}{3a^2})^2=\frac{(m+n)^2}{(3a^2)^2}=\frac{(m+n)^2}{9a^4}$.

6 填空:
(1)$(\frac {-2x^{2}}{y})^{3}= $____;
(2)$(\frac {5x^{3}y}{-3z^{4}})^{2}= $____.

答案：
(1)$-\frac{8x^6}{y^3}$;
(2)$\frac{25x^6y^2}{9z^8}$
(1)$(-\frac{2x^2}{y})^3=-(\frac{2x^2}{y})^3=-\frac{8x^6}{y^3}$.
(2)$(\frac{5x^3y}{-3z^4})^2=\frac{25x^6y^2}{9z^8}$.

7 [2023河北中考]化简$x^{3}(\frac {y^{3}}{x})^{2}$的结果是( )

A.$xy^{6}$
B.$xy^{5}$
C.$x^{2}y^{5}$
D.$x^{2}y^{6}$

答案： A $x^3(\frac{y^3}{x})^2=x^3\cdot\frac{y^6}{x^2}=xy^6$.

8 计算$(\frac {a-b}{b})^{2}\cdot \frac {b}{a^{2}-b^{2}}$的结果是( )

A.$\frac {1}{b}$
B.$\frac {a-b}{ab+b^{2}}$
C.$\frac {a-b}{a+b}$
D.$\frac {1}{b(a+b)}$

答案： B $(\frac{a-b}{b})^2\cdot\frac{b}{a^2-b^2}=\frac{(a-b)^2}{b^2}\cdot\frac{b}{(a+b)(a-b)}=$ $\frac{a-b}{b(a+b)}=\frac{a-b}{ab+b^2}$.

9 [2025北京房山区期末]计算:$(\frac {3n}{2m})^{2}(-\frac {m}{n})^{3}= $____.

答案： $-\frac{9m}{4n}$ $(\frac{3n}{2m})^2(-\frac{m}{n})^3=\frac{9n^2}{4m^2}\cdot(-\frac{m^3}{n^3})=-\frac{9n^2}{4m^2}\cdot\frac{m^3}{n^3}=$ $-\frac{9m}{4n}$.

10 若$\frac {x}{y}= 3$,则$\frac {1}{x^{2}-y^{2}}÷(\frac {1}{x-y})^{2}$的值是____.

答案： $\frac{1}{2}$ $\frac{1}{x^2-y^2}÷(\frac{1}{x-y})^2=\frac{1}{(x+y)(x-y)}\cdot(x-y)^2=\frac{x-y}{x+y}$. 因为$\frac{x}{y}=3$,所以$x=3y$,所以原式$=\frac{3y-y}{3y+y}=\frac{1}{2}$.

11 教材P151T4变式计算:
(1)$(-\frac {2a}{b^{2}})^{3}\cdot (\frac {2b}{a})^{2}÷(-\frac {2b}{a})^{2};$
(2)$4a^{2}b÷(-\frac {a}{2b})^{2}\cdot (-\frac {b}{8a});$
(3)$(\frac {x-4}{x-1})^{2}÷\frac {x^{2}-8x+16}{x-1}\cdot \frac {x^{2}-1}{x+2}.$

答案： 解:
(1)$(-\frac{2a}{b^2})^3\cdot(\frac{2b}{a})^2÷(-\frac{2b}{a})^2$ $=-\frac{8a^3}{b^6}\cdot\frac{4b^2}{a^2}÷\frac{4b^2}{a^2}$ $=-\frac{8a^3}{b^6}\cdot\frac{4b^2}{a^2}\cdot\frac{a^2}{4b^2}$ $=-\frac{8a^3}{b^6}$.
(2)$4a^2b÷(-\frac{a}{2b})^2\cdot(-\frac{b}{8a})$ $=4a^2b÷\frac{a^2}{4b^2}\cdot(-\frac{b}{8a})$ $=4a^2b\cdot\frac{4b^2}{a^2}\cdot(-\frac{b}{8a})$ $=-\frac{2b^4}{a}$.
(3)$(\frac{x-4}{x-1})^2÷\frac{x^2-8x+16}{x-1}\cdot\frac{x^2-1}{x+2}$ $=\frac{(x-4)^2}{(x-1)^2}\cdot\frac{x-1}{(x-4)^2}\cdot\frac{(x+1)(x-1)}{x+2}$ $=\frac{x+1}{x+2}$.

12 先化简,再求值:$(\frac {2ab^{2}}{a+b})^{3}÷(\frac {ab^{3}}{a^{2}-b^{2}})^{2}\cdot [\frac {1}{2(a-b)}]^{2}$,其中$a= -\frac {1}{2},b= \frac {2}{3}.$

答案： 解:$(\frac{2ab^2}{a+b})^3÷(\frac{ab^3}{a^2-b^2})^2\cdot[\frac{1}{2(a-b)}]^2$ $=\frac{(2ab^2)^3}{(a+b)^3}\cdot\frac{(a^2-b^2)^2}{(ab^3)^2}\cdot\frac{1}{4(a-b)^2}$ $=\frac{8a^3b^6}{(a+b)^3}\cdot\frac{(a+b)^2(a-b)^2}{a^2b^6}\cdot\frac{1}{4(a-b)^2}$ $=\frac{2a}{a+b}$. 当$a= -\frac{1}{2},b= \frac{2}{3}$时,原式$=\frac{2×(-\frac{1}{2})}{-\frac{1}{2}+\frac{2}{3}}=-6$.