Rút gọn phân thức đại số sau: \(\left( {\dfrac{{x + 1}}{{2x - 2}} + \dfrac{3}{{{x^2} - 1}} - \dfrac{{x + 3}}{{2x + 2}}} \right)\)\(.\dfrac{{4{x^2} - 4}}{5}\)

* Hướng dẫn giải

\(\displaystyle  \left( {{{x + 1} \over {2x - 2}} + {3 \over {{x^2} - 1}} - {{x + 3} \over {2x + 2}}} \right)\)\(\displaystyle .{{4{x^2} - 4} \over 5} \) 
\(\displaystyle = {{{{\left( {x + 1} \right)}^2} + 3.2 - \left( {x + 3} \right)\left( {x - 1} \right)} \over {2\left( {x - 1} \right)\left( {x + 1} \right)}}\)\(\displaystyle .{{4\left( {x - 1} \right)\left( {x + 1} \right)} \over 5} \) 
\(\displaystyle = {{{x^2} + 2x + 1 + 6 - \left( {{x^2} - x + 3x - 3} \right)} \over {2\left( {x - 1} \right)\left( {x + 1} \right)}}\)\(\displaystyle .{{4\left( {x - 1} \right)\left( {x + 1} \right)} \over 5} \) 
\(\displaystyle   = {{10} \over {2\left( {x - 1} \right)\left( {x + 1} \right)}}\)\(\displaystyle .{{4\left( {x - 1} \right)\left( {x + 1} \right)} \over 5} = 4  \)