Bài 1. Ta có:\( {x^2} - x = x\left( {x - 1} \right);\)
\({x^2} - 1 = \left( {x - 1} \right)\left( {x + 1} \right);\)
\({x^2} + 2x + 1 = {\left( {x + 1} \right)^2}\)
\( MTC = x\left( {x - 1} \right){\left( {x + 1} \right)^2}\)
Vậy: \( {1 \over {{x^2} - x}} = {{{{\left( {x + 1} \right)}^2}} \over {x\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}};\)
\({3 \over {{x^2} - 1}} = {{3x\left( {x + 1} \right)} \over {x\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}};\)
\( {2 \over {{x^2} + 2x + 1}} = {{2x\left( {x - 1} \right)} \over {x\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}\) .
Bài 2. Ta có:\( {1 \over {1 - 9{x^2}}} = {{ - 1} \over {9{x^2} - 1}}\)
\( MTC = 9{x^2} - 1 = \left( {3x - 1} \right)\left( {3x + 1} \right)\)
Vậy: \( {x \over {3x - 1}} = {{x\left( {3x - 1} \right)} \over {9{x^2} - 1}};{{2x - 1} \over {3x + 1}} = {{\left( {2x - 1} \right)\left( {3x - 1} \right)} \over {9{x^2} - 1}}\)
