Bài 1. Ta có: \( {{5x - 6} \over {4 - {x^2}}} = {{ - \left( {5x - 6} \right)} \over { - \left( {4 - {x^2}} \right)}} = {{\left( {6 - 5x} \right)} \over {{x^2} - 4}}\)
\( MTC = {x^2} - 4 = \left( {x - 2} \right)\left( {x + 2} \right)\)
Bài 2.
a) Ta có: \( MTC = {a^3} - 27 \)\(\;= \left( {a - 3} \right)\left( {{a^2} + 3a + 9} \right)\)
Vậy: \( {{a - 3} \over {{a^2} + 3a + 9}} = {{{{\left( {a - 3} \right)}^2}} \over {\left( {a - 3} \right)\left( {{a^2} + 3a + 9} \right)}} = {{{{\left( {a - 3} \right)}^2}} \over {{a^3} - 27}};\)
\( {1 \over {a - 3}} = {{{a^3} + 3a + 9} \over {\left( {a - 3} \right)\left( {{a^2} + 3a + 9} \right)}} = {{{a^2} + 3a + 9} \over {{a^3} - 27}}\)
b) \( MTC = 12{x^2}{y^3}\)
Vậy: \( {2 \over {3{y^3}}} = {{8{x^2}} \over {12{x^2}{y^2}}};\)
\({1 \over {6{x^2}y}} = {{2{y^2}} \over {12{x^2}{y^3}}};{5 \over {12x{y^2}}} = {{5xy} \over {12x{y^3}}}\)
