Bài 1. Ta có : \({{\left( {x - 2} \right)m} \over {ym}} = {{\left( {x - 2} \right)\left( {x + 2} \right)} \over {y\left( {x + 2} \right)}}\)
Vậy \(m = x + 2.\)
Bài 2. Ta có : \({{{x^2}\left( {x - 1} \right)} \over {x{{\left( {1 - x} \right)}^2}}} = {{{x^2}\left( {x - 1} \right)} \over {x{{\left( {x - 1} \right)}^2}}} = {x \over {x - 1}}\)(đpcm).
Bài 3.
a) Ta có :\({{3x + 2} \over {{{\left( {x - 1} \right)}^2}}} = {{\left( {3x + 2} \right)\left( {x + 1} \right)} \over {{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}};\)
\({1 \over {{x^2} - 1}} = {1 \over {\left( {x - 1} \right)\left( {x + 1} \right)}} = {{x - 1} \over {{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}.\)
b) Ta có : \({{x + 1} \over {x - 1}} = {{ - \left( {x + 1} \right)} \over { - \left( {x - 1} \right)}} = {{ - \left( {x + 1} \right)} \over {1 - x}} = {{ - \left( {x + 1} \right)\left( {1 + x} \right)} \over {\left( {1 - x} \right)\left( {1 + x} \right)}} \)\(\;= {{ - {{\left( {x + 1} \right)}^2}} \over {1 - {x^2}}}\)
