Đáp án:
a. \(\left[ \begin{array}{l}
x = - \dfrac{7}{5}\\
x = - \dfrac{5}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne - 1\\
{\left[ {{{\left( {\dfrac{{2x + 3}}{{ - x - 1}}} \right)}^5}} \right]^2} = {\left[ {{{\left( { - \dfrac{1}{2}} \right)}^2}} \right]^5}\\
\to {\left( {\dfrac{{2x + 3}}{{ - x - 1}}} \right)^{10}} = {\left( { - \dfrac{1}{2}} \right)^{10}}\\
\to \left| {\dfrac{{2x + 3}}{{ - x - 1}}} \right| = \left| { - \dfrac{1}{2}} \right|\\
\to \left| {\dfrac{{2x + 3}}{{ - x - 1}}} \right| = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
\dfrac{{2x + 3}}{{ - x - 1}} = \dfrac{1}{2}\\
\dfrac{{2x + 3}}{{ - x - 1}} = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x + 6 = - x - 1\\
4x + 6 = x + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = - 7\\
3x = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{7}{5}\\
x = - \dfrac{5}{3}
\end{array} \right.\left( {TM} \right)\\
b.DK:x \ne \dfrac{5}{3}\\
{\left[ {{{\left( {\dfrac{{ - x + 3}}{{ - 5 + 3x}}} \right)}^5}} \right]^2} = {\left[ {{{\left( { - \dfrac{2}{3}} \right)}^4}} \right]^5}\\
\to {\left( {\dfrac{{ - x + 3}}{{ - 5 + 3x}}} \right)^{10}} = {\left( { - \dfrac{2}{3}} \right)^{20}}\\
\to \left| {\dfrac{{ - x + 3}}{{ - 5 + 3x}}} \right| = {\left( { - \dfrac{2}{3}} \right)^2}\\
\to \left| {\dfrac{{ - x + 3}}{{ - 5 + 3x}}} \right| = \dfrac{4}{9}\\
\to \left[ \begin{array}{l}
\dfrac{{ - x + 3}}{{ - 5 + 3x}} = \dfrac{4}{9}\\
\dfrac{{ - x + 3}}{{ - 5 + 3x}} = - \dfrac{4}{9}
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 9x + 27 = - 20 + 12x\\
- 9x + 27 = 20 - 12x
\end{array} \right.\\
\to \left[ \begin{array}{l}
21x = - 47\\
3x = - 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{{47}}{{21}}\\
x = - \dfrac{7}{3}
\end{array} \right.\left( {TM} \right)
\end{array}\)
