Bài 1. Ta có: \( 4x - 4 = 4\left( {x - 1} \right);\)
\({x^2} - 1 = \left( {x - 1} \right)\left( {x + 1} \right);\)
\( 3{x^2} + 3x = 3x\left( {x + 1} \right)\)
MTC: \( 12x\left( {x - 1} \right)\left( {x + 1} \right)\)
Bài 2.
a) \( MTC = 9{x^2} - 1 = \left( {3x - 1} \right)\left( {3x + 1} \right)\)
Ta có: \( {{4x} \over {1 - 3x}} = {{ - 4x} \over {3x - 1}} = {{ - 4x\left( {3x + 1} \right)} \over {\left( {3x - 1} \right)\left( {3x + 1} \right)}}\)\(\; = {{ - 4x\left( {3x + 1} \right)} \over {9{x^2} - 1}}\)
b) Ta có: \( {x^3} + 8 = \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\)
\( MTC = {x^3} + 8\)
Vậy: \( {3 \over {x + 2}} = {{3\left( {{x^2} - 2x + 4} \right)} \over {\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}} = {{3\left( {{x^2} - 2x + 4} \right)} \over {{x^3} + 8}}\)
\( {{x + 2} \over {{x^2} - 2x + 4}} = {{{{\left( {x + 2} \right)}^2}} \over {\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}} = {{{{\left( {x + 2} \right)}^2}} \over {{x^3} + 8}}\)
