Question

6．计算：（$\frac{{b}^{2}}{{a}^{2}}$-$\frac{{a}^{2}}{{b}^{2}}$）（$\frac{{b}^{2}}{{a}^{2}}$+$\frac{{a}^{2}}{{b}^{2}}$-1）（$\frac{{b}^{2}}{{a}^{2}}$+$\frac{{a}^{2}}{{b}^{2}}$+1）．

Answer 1

分析 先算括号里面的，再算乘法即可．

解答 解：原式=$\frac{{b}^{4}-{a}^{4}}{{a}^{2}{b}^{2}}$•[$\frac{{b}^{4}+{a}^{4}}{{a}^{2}{b}^{2}}$-1]•[$\frac{{b}^{4}+{a}^{4}}{{a}^{2}{b}^{2}}$+1]
=$\frac{{b}^{4}-{a}^{4}}{{a}^{2}{b}^{2}}$•[（$\frac{{b}^{4}+{a}^{4}}{{a}^{2}{b}^{2}}$）²-1]
=$\frac{{b}^{4}-{a}^{4}}{{a}^{2}{b}^{2}}$•$\frac{{（a}^{4}+{b}^{4}）^{2}}{{a}^{4}{b}^{4}}$-$\frac{{b}^{4}-{a}^{4}}{{a}^{2}{b}^{2}}$
=$\frac{{{（a}^{4}+{b}^{4}）}^{2}{（b}^{4}-{a}^{4}）}{{a}^{6}{b}^{6}}$-$\frac{（{b}^{4}-{a}^{4}）•{a}^{4}{b}^{4}}{{a}^{6}{b}^{6}}$
=$\frac{{{（a}^{4}+{b}^{4}）}^{\;}{（b}^{8}-{a}^{8}）}{{a}^{6}{b}^{6}}$-$\frac{（{b}^{4}-{a}^{4}）•{a}^{4}{b}^{4}}{{a}^{6}{b}^{6}}$
=$\frac{{a}^{4}{b}^{8}-{a}^{12}+{b}^{12}-{a}^{8}{b}^{4}-{a}^{4}{b}^{8}+{a}^{8}{b}^{4}}{{a}^{6}{b}^{6}}$
=$\frac{{b}^{12}-{a}^{12}}{{a}^{6}{b}^{6}}$．

点评 本题考查的是分式的混合运算，熟知分式混合运算的法则是解答此题的关键．