Ta có \({x^2} - 9 = \left( {x + 3} \right)\left( {x - 3} \right)\)
\(\displaystyle {{3x} \over {x + 3}} = {{3x\left( {x - 3} \right)} \over {\left( {x + 3} \right)\left( {x - 3} \right)}} \)\(\,\displaystyle= {{3{x^2} - 9x} \over {{x^2} - 9}}\)
\(\displaystyle{{x - 1} \over {x - 3}} = {{\left( {x - 1} \right)\left( {x + 3} \right)} \over {\left( {x - 3} \right)\left( {x + 3} \right)}}\)\(\,\displaystyle = {{{x^2} + 2x - 3} \over {{x^2} - 9}} \)
\(\displaystyle {x^2} + 9 = {{\left( {{x^2} + 9} \right)\left( {{x^2} - 9} \right)} \over {{x^2} - 9}} \)\(\,\displaystyle = {{{x^4} - 81} \over {{x^2} - 9}} \)
