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