Bài 1.
a) \({{{a^2} - {b^2}} \over {9{b^2}}}:{{a + b} \over {3b}} = {{3b\left( {{a^2} - {b^2}} \right)} \over {9{b^2}\left( {a + b} \right)}} = {{a - b} \over {3b}}.\)
b) \({{{a^2} - 2a + 1} \over {2a + 1}}:{{a - 1} \over {4{a^2} - 1}} \)
\(= {{{{\left( {a - 1} \right)}^2}.\left( {4{a^2} - 1} \right)} \over {\left( {2a + 1} \right)\left( {a - 1} \right)}} \)
\(= \left( {a - 1} \right)\left( {2a - 1} \right).\)
Bài 2.
\({{a - 1} \over a}:\left( {{{{a^2} + 1} \over {{a^2} + 2a}} - {2 \over {a + 2}}} \right) \)
\(= {{a - 1} \over a}:\left[ {{{{a^2} + 1 - 2a} \over {a\left( {a + 2} \right)}}} \right] \)
\(= {{a - 1} \over a}.{{a\left( {a + 2} \right)} \over {{{\left( {a - 1} \right)}^2}}} = {{a + 2} \over {a - 1}}.\)
Bài 3.
\(P = {{{a^2}b - 4{b^3}} \over {3a{b^2}}}:{{{a^2} - 2ab} \over {{a^2}b}} \)
\(\;\;\;\;= {{b\left( {{a^2} - 4{b^2}} \right)} \over {3a{b^2}}}.{{{a^2}b} \over {{a^2} - 2ab}}\)
\(\;\;\;\; = {{{a^2}{b^2}\left( {a - 2b} \right)\left( {a + 2b} \right)} \over {3{a^2}{b^2}\left( {a - 2b} \right)}} = {{a + 2b} \over 3}.\)
