Đáp án:
3) \(\left[ \begin{array}{l}
x = 0\\
x = \dfrac{{31}}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne - 2\\
\dfrac{{x + 3}}{{x + 2}} = \dfrac{5}{4}\\
\to 4x + 12 = 5x + 10\\
\to x = 2\\
2)DK:x \ne \left\{ { - 4; - 1} \right\}\\
\dfrac{{2x + 3}}{{x + 4}} = \dfrac{{2x + 5}}{{x + 1}}\\
\to \left( {2x + 3} \right)\left( {x + 1} \right) = \left( {2x + 5} \right)\left( {x + 4} \right)\\
\to 2{x^2} + 2x + 3x + 3 = 2{x^2} + 8x + 5x + 20\\
\to 8x = - 17\\
\to x = - \dfrac{{17}}{8}\\
3)DK:x \ne \left\{ { - 3;5} \right\}\\
\dfrac{{4x - 3}}{{x + 3}} = \dfrac{{x + 5}}{{x - 5}}\\
\to \left( {4x - 3} \right)\left( {x - 5} \right) = \left( {x + 5} \right)\left( {x + 3} \right)\\
\to 4{x^2} - 20x - 3x + 15 = {x^2} + 8x + 15\\
\to 3{x^2} - 31x = 0\\
\to x\left( {3x - 31} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{31}}{3}
\end{array} \right.
\end{array}\)
