Đáp án:
b) \(\left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = \dfrac{4}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - \dfrac{5}{4}; - \dfrac{1}{2}} \right\}\\
\dfrac{{3x - 5}}{{2x + 1}} = \dfrac{{4 - 3x}}{{4x + 5}}\\
\to \left( {3x - 5} \right)\left( {4x + 5} \right) = \left( {4 - 3x} \right)\left( {2x + 1} \right)\\
\to 12{x^2} - 5x - 25 = - 6{x^2} + 5x + 4\\
\to 18{x^2} - 10x - 29 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{5 + \sqrt {547} }}{{18}}\\
x = \dfrac{{5 - \sqrt {547} }}{{18}}
\end{array} \right.\\
b)DK:x \ne \left\{ {\dfrac{2}{3};2} \right\}\\
\dfrac{{3x + 7}}{{6x - 4}} - \dfrac{{5x - 2}}{{4 - 2x}} = 1\\
\to \dfrac{{\left( {3x + 7} \right)\left( {4 - 2x} \right) - \left( {5 - 2x} \right)\left( {6x - 4} \right)}}{{\left( {6x - 4} \right)\left( {4 - 2x} \right)}} = \dfrac{{\left( {6x - 4} \right)\left( {4 - 2x} \right)}}{{\left( {6x - 4} \right)\left( {4 - 2x} \right)}}\\
\to - 6{x^2} - 14x + 12x + 28 + 12{x^2} - 38x + 20 = - 12{x^2} + 32x - 16\\
\to 18{x^2} - 72x + 64 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = \dfrac{4}{3}
\end{array} \right.\\
c)DK:x \ne \left\{ { - \dfrac{5}{2}; - \dfrac{3}{2}} \right\}\\
\dfrac{{3x - 1}}{{2x + 5}} = \dfrac{{1 - 6x}}{{2x + 3}}\\
\to \left( {3x - 1} \right)\left( {2x + 3} \right) - \left( {2x + 5} \right)\left( {1 - 6x} \right) = 0\\
\to 6{x^2} + 7x - 3 + 12{x^2} + 28x - 5 = 0\\
\to 18{x^2} + 35x - 8 = 0\\
\to \left[ \begin{array}{l}
x = 0,2066163986\\
x = - 2,151060843
\end{array} \right.
\end{array}\)
