Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne - 1\\
x.\left( {\dfrac{{3 - x}}{{x + 1}}} \right).\left( {x + \dfrac{{3 - x}}{{x + 1}}} \right) = 2\\
\to x.\left( {\dfrac{{3 - x}}{{x + 1}}} \right).\left( {\dfrac{{x + 1 + 3 - x}}{{x + 1}}} \right) = 2\\
\to \dfrac{{3x - {x^2}}}{{x + 1}}.\dfrac{4}{{x + 1}} = 2\\
\to \dfrac{{12x - 4{x^2}}}{{{x^2} + 2x + 1}} = 2\\
\to 12x - 4{x^2} = 2{x^2} + 4x + 2\\
\to 6{x^2} - 8x + 2 = 0\\
\to 2\left( {x - 1} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{3}
\end{array} \right.\left( {TM} \right)
\end{array}\)
