《2025年自我提升与评价八年级数学上册人教版》

7. 计算:$\left(-x^3y^2\right)^2\cdot\frac{x}{y^3}=$$x^7y$.

答案： x^7y

8. 计算:$\left(\frac{a-b}{ab}\right)^3\cdot\left(\frac{ab}{a-b}\right)^2÷(a-b)^2= $$\boxed{\dfrac{1}{ab(a-b)}}$.

答案：
$\boxed{\dfrac{1}{ab(a-b)}}$

9. 计算:
(1)$\left(\frac{b}{2a}\right)^2÷\left(\frac{-b}{a}\right)\cdot\left(-\frac{3b}{4a}\right)^3$;
(2)$\left(\frac{a^2b}{-c}\right)^3\cdot\left(\frac{c^2}{-ab}\right)^2÷\left(\frac{bc}{a}\right)^4$;
(3)$\frac{m^2-36}{m^2+10m+25}÷\frac{m-6}{2m+10}\cdot\frac{m+5}{m^2+6m}$;
(4)$\left(\frac{x^2-y^2}{xy}\right)^2÷(x+y)\cdot\left(\frac{x}{x-y}\right)^2$.

答案：
(1) 原式$=\frac{b^{2}}{4a^{2}}÷\left(-\frac{b}{a}\right)\cdot\left(-\frac{27b^{3}}{64a^{3}}\right)$
$=\frac{b^{2}}{4a^{2}}\cdot\left(-\frac{a}{b}\right)\cdot\left(-\frac{27b^{3}}{64a^{3}}\right)$
$=\frac{b^{2}\cdot a\cdot27b^{3}}{4a^{2}\cdot b\cdot64a^{3}}$
$=\frac{27b^{4}}{256a^{4}}$
(2) 原式$=-\frac{a^{6}b^{3}}{c^{3}}\cdot\frac{c^{4}}{a^{2}b^{2}}÷\frac{b^{4}c^{4}}{a^{4}}$
$=-\frac{a^{6}b^{3}}{c^{3}}\cdot\frac{c^{4}}{a^{2}b^{2}}\cdot\frac{a^{4}}{b^{4}c^{4}}$
$=-\frac{a^{6}b^{3}\cdot c^{4}\cdot a^{4}}{c^{3}\cdot a^{2}b^{2}\cdot b^{4}c^{4}}$
$=-\frac{a^{8}}{b^{3}c^{3}}$
(3) 原式$=\frac{(m-6)(m+6)}{(m+5)^{2}}÷\frac{m-6}{2(m+5)}\cdot\frac{m+5}{m(m+6)}$
$=\frac{(m-6)(m+6)}{(m+5)^{2}}\cdot\frac{2(m+5)}{m-6}\cdot\frac{m+5}{m(m+6)}$
$=\frac{(m-6)(m+6)\cdot2(m+5)\cdot(m+5)}{(m+5)^{2}\cdot(m-6)\cdot m(m+6)}$
$=\frac{2}{m}$
(4) 原式$=\frac{(x-y)^{2}(x+y)^{2}}{x^{2}y^{2}}÷(x+y)\cdot\frac{x^{2}}{(x-y)^{2}}$
$=\frac{(x-y)^{2}(x+y)^{2}}{x^{2}y^{2}}\cdot\frac{1}{x+y}\cdot\frac{x^{2}}{(x-y)^{2}}$
$=\frac{(x-y)^{2}(x+y)^{2}\cdot x^{2}}{x^{2}y^{2}\cdot(x+y)\cdot(x-y)^{2}}$
$=\frac{x+y}{y^{2}}$