Số nghiệm của phương trình $\dfrac{{{x^2} + 3x + 2}}{{x + 3}} - \dfrac{{{x^2} + 2x + 1}}{{x - 1}} = \dfrac{{4x + 4}}{{{x^2} + 2x - 3}}$ là

Ta có  $\dfrac{{{x^2} + 3x + 2}}{{x + 3}} - \dfrac{{{x^2} + 2x + 1}}{{x - 1}} = \dfrac{{4x + 4}}{{{x^2} + 2x - 3}}$

\( \Leftrightarrow \dfrac{{{x^2} + 2x + x + 2}}{{x + 3}} - \dfrac{{{{\left( {x + 1} \right)}^2}}}{{x - 1}} = \dfrac{{4\left( {x + 1} \right)}}{{{x^2} - x + 3x - 3}}\)

\( \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x + 1} \right)}}{{x + 3}} - \dfrac{{{{\left( {x + 1} \right)}^2}}}{{x - 1}} = \dfrac{{4\left( {x + 1} \right)}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}\)

ĐK: \(x \ne \left\{ {1; - 3} \right\}\)

Khi đó \(pt \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x + 1} \right)\left( {x - 1} \right) - {{\left( {x + 1} \right)}^2}\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 1} \right)}} = \dfrac{{4\left( {x + 1} \right)}}{{\left( {x + 3} \right)\left( {x - 1} \right)}}\)

\(\)\( \Rightarrow \left( {x + 2} \right)\left( {x + 1} \right)\left( {x - 1} \right) - {\left( {x + 1} \right)^2}\left( {x + 3} \right) - 4\left( {x + 1} \right) = 0\)

\( \Leftrightarrow \left( {x + 1} \right)\left[ {\left( {x + 2} \right)\left( {x - 1} \right) - \left( {x + 1} \right)\left( {x + 3} \right) - 4} \right] = 0\)

\( \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + x - 2 - {x^2} - 4x - 3 - 4} \right) = 0\)

\(\begin{array}{l} \Leftrightarrow \left( {x + 1} \right)\left( { - 3x - 9} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 1 = 0\\ - 3x - 9 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 1\left( {TM} \right)\\x =  - 3\,\left( {KTM} \right)\end{array} \right.\end{array}\)

Vậy phương trình có nghiệm duy nhất \(x =  - 1\).