Đáp án:
d. \(x = - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne - 5\\
\dfrac{{x - 3}}{{x + 5}} = \dfrac{5}{7}\\
\to 7x - 21 = 5x + 25\\
\to 2x = 46\\
\to x = 23\\
2)DK:x \ne 1\\
\dfrac{7}{{x - 1}} = \dfrac{{x + 1}}{9}\\
\to \left( {x - 1} \right)\left( {x + 1} \right) = 63\\
\to {x^2} - 1 = 63\\
\to {x^2} = 64\\
\to \left| x \right| = 8\\
\to \left[ \begin{array}{l}
x = 8\\
x = - 8
\end{array} \right.\\
3)DK:x \ne - 4\\
\dfrac{{x + 4}}{{20}} = \dfrac{5}{{x + 4}}\\
\to {\left( {x + 4} \right)^2} = 100\\
\to \left| {x + 4} \right| = 10\\
\to \left[ \begin{array}{l}
x + 4 = 10\\
x + 4 = - 10
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
x = - 14
\end{array} \right.\\
4)DK:x \ne \left\{ { - 3; - 2} \right\}\\
\dfrac{{x - 1}}{{x + 2}} = \dfrac{{x - 2}}{{x + 3}}\\
\to \left( {x - 1} \right)\left( {x + 3} \right) = \left( {x + 2} \right)\left( {x - 2} \right)\\
\to {x^2} + 2x - 3 = {x^2} - 4\\
\to 2x = - 1\\
\to x = - \dfrac{1}{2}
\end{array}\)
