a) \(\displaystyle {{{x^2} + 3x + 2} \over {3x + 6}}\)
\(\displaystyle = {{{x^2} + x + 2x + 2} \over {3\left( {x + 2} \right)}} \)
\(\displaystyle= {{x\left( {x + 1} \right) + 2\left( {x + 1} \right)} \over {3\left( {x + 2} \right)}}\)
\(\displaystyle= {{\left( {x + 1} \right)\left( {x + 2} \right)} \over {3\left( {x + 2} \right)}} = {{x + 1} \over 3}\)
\(\displaystyle {{2{x^2} + x - 1} \over {6x - 3}}\)
\(\displaystyle = {{2{x^2} + 2x - x - 1} \over {3\left( {2x - 1} \right)}} \)
\(\displaystyle = {{2x\left( {x + 1} \right) - \left( {x + 1} \right)} \over {3\left( {2x - 1} \right)}} \)
\(\displaystyle = {{\left( {x + 1} \right)\left( {2x - 1} \right)} \over {3\left( {2x - 1} \right)}} = {{x +1} \over 3}\)
Vậy \(\displaystyle {{{x^2} + 3x + 2} \over {3x + 6}}= {{2{x^2} + x - 1} \over {6x - 3}}\)
b) \(\displaystyle {{15x - 10} \over {3{x^2} + 3x - \left( {2x + 2} \right)}}\)
\(\displaystyle = {{5\left( {3x - 2} \right)} \over {3x\left( {x + 1} \right) - 2\left( {x + 1} \right)}}\)
\(\displaystyle = {{5\left( {3x - 2} \right)} \over {\left( {x + 1} \right)\left( {3x - 2} \right)}} = {5 \over {x + 1}}\)
\(\displaystyle {{5{x^2} - 5x + 5} \over {{x^3} + 1}}\) \(\displaystyle = {{5\left( {{x^2} - x + 1} \right)} \over {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\)\(\,\displaystyle = {5 \over {x + 1}}\)
Vậy \(\displaystyle {{15x - 10} \over {3{x^2} + 3x - \left( {2x + 2} \right)}}\)\(\,\displaystyle = {{5{x^2} - 5x + 5} \over {{x^3} + 1}}\)
