Giải phương trình sau: \(\left( {2x - 1} \right)\left( {x + 5} \right) = 2\left( {{x^2} + \dfrac{3}{2}} \right) - 7x\)

* Hướng dẫn giải

Điều kiện xác định: \(\left\{ \begin{array}{l}x - 2 \ne 0\\x + 2 \ne 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x \ne 2\\x \ne  - 2\end{array} \right.\)

\(\begin{array}{l}B = \dfrac{{6 - 7x}}{{{x^2} - 4}} + \dfrac{3}{{x + 2}} - \dfrac{2}{{2 - x}}\\\,\,\,\,\, = \dfrac{{6 - 7x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{3}{{x + 2}} + \dfrac{2}{{x - 2}}\\\,\,\,\,\, = \dfrac{{6 - 7x + 3\left( {x - 2} \right) + 2\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\\,\,\,\,\, = \dfrac{{6 - 7x + 3x - 6 + 2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\\,\,\,\, = \dfrac{{ - 2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\\,\,\,\, = \dfrac{{ - 2\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\\,\,\,\, = \dfrac{{ - 2}}{{x + 2}}\end{array}\)