第122页 - 启东中学作业本八年级数学人教版

$解:原式=\frac{(x-2)(x-3)}{(x+4)(x-4)}×\frac {4-x}{x-3}×\frac{(x+1)(x+4)}{(2+x)(2-x)}$

$=\frac {x+1}{x+2}$

$解:原式=\frac{16x^{2}}{y^{5}}$

①

运算顺序错误

④

a等于1时，原式无意义

$解:原式=\frac{(a+b)^{3}}{8a^{3}b^{6}}÷\frac{(a+b)^{2}(a-b)^{2}}{a^{2}b^{6}}÷\frac{1}{4(a-b)^{2}}=\frac{(a+b)^{3}}{8a^{3}b^{6}}×\frac{a^{2}b^{6}}{(a+b)^{2}(a-b)^{2}}×4(a-b)^{2}=\frac{a+b}{2a}$

$当a=-\frac{1}{2},b=\frac{2}{3}时，\frac{a+b}{2a}= \frac {-\frac {1}{2}+\frac {2}{3}}{2×(-\frac {1}{2})}=-\frac {1}{6}$

$解:原式=\frac {2x+y}{(x-y)^{2}}×(x-y)=\frac {2x+y}{x-y}$

$∵x-3y=0，∴x=3y$

$原式=\frac {7y}{2y}=\frac{7}{2}$

$解:(a-1)÷\frac{a^{2}-1}{a+1}×\frac{a+1}{ab^{2}}=(a-1)×\frac{a+1}{(a+1)(a-1)}×\frac{a+1}{ab^{2}}=\frac {a+1}{ab^{2}}\ $

$当a=2,b=2时,原式=\frac{3}{8}$

$解:原式=a(a-\frac {1}{a})(a+\frac{1}{a})(a^{2}+\frac{1}{a^{2}})(a^{4}+\frac{1}{a^{4}})×(a^{8}+\frac{1}{a^{8}})$

$=a(a^{2}-\frac{1}{a^{2}})(a^{2}+\frac{1}{a^{2}})(a^{4}+\frac{1}{a^{4}})(a^{8}+\frac{1}{a^{8}})$

$=a(a^{4}-\frac{1}{a^{4}})(a^{4}+\frac{1}{a^{4}})(a^{8}+\frac{1}{a^{8}})=a(a^{8}-\frac{1}{a^{8}})(a^{8}+\frac{1}{a^{8}})$

$=a(a^{16}-\frac{1}{a^{16}}) =a^{17}-\frac{1}{a^{15}}$