Điều kiện: \(x \ne -\dfrac{1}{2}\) và \(x \ne -\dfrac{7}{2}\)
\(\begin{array}{l} \dfrac{{\left( {2x + 3} \right)\left( {2x + 7} \right)}}{{\left( {2x + 1} \right)\left( {2x + 7} \right)}} - \dfrac{{\left( {2x + 5} \right)\left( {2x + 1} \right)}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} - \dfrac{{6{x^2} + 9x - 9}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Leftrightarrow \dfrac{{4{x^2} + 20x + 21 - 4{x^2} - 12x - 5}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{4{x^2} + 16x + 7 - 6{x^2} - 9x + 9}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Leftrightarrow \dfrac{{8x + 16}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{ - 2{x^2} + 7x + 16}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Rightarrow 8x + 16 = - 2{x^2} + 7x + 16 \Leftrightarrow 2{x^2} + x = 0 \Leftrightarrow x\left( {2x + 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0 \text{(nhận)}\\ x = - \dfrac{1}{2} \text{(loại)} \end{array} \right. \end{array}\)
Vậy phương trình có nghiệm duy nhất $x=0$
