Giải phương trình sau: \({\frac{{x + 2}}{{x + 3}} + \frac{{2x - 1}}{{x - 3}} = \frac{{13x - 9}}{{{x^2} - 9}}}\)

* Hướng dẫn giải

\(\,\,\dfrac{{x + 2}}{{x + 3}} + \dfrac{{2x - 1}}{{x - 3}} = \dfrac{{13x - 9}}{{{x^2} - 9}}\)    Điều kiện: \(x \ne 3;\,\,x \ne  - 3.\)

\( \Leftrightarrow \,\,\dfrac{{x + 2}}{{x + 3}} + \dfrac{{2x - 1}}{{x - 3}} = \dfrac{{13x - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\)

\( \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right) + \left( {2x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\)\( = \dfrac{{13x - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\)

\( \Leftrightarrow \left( {x + 2} \right)\left( {x - 3} \right) + \left( {2x - 1} \right)\left( {x + 3} \right)\)\( = 13x - 9\)

\(\begin{array}{l} \Leftrightarrow {x^2} - x - 6 + 2{x^2} + 5x - 3 = 13x - 9\\ \Leftrightarrow 3{x^2} - 9x = 0\\ \Leftrightarrow 3x\left( {x - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}3x = 0\\x - 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\,\,\,\left( {tm} \right)\\x = 3\,\,\left( {ktm} \right)\end{array} \right.\end{array}\)

Vậy phương trình có nghiệm là \(x = 0\)