(-3 - $\frac{3}{x}$ - $\frac{1}{3}$ ) : ( 1+ $\frac{2}{5}$ + $\frac{2}{3}$ ) = $\frac{-5}{4}$ ( 4x -1 ) ² = ( 1- 4x ) ^ 4 Mọi người giúp với!!

Đáp án:

$\begin{array}{l}
\left( { - 3 - \dfrac{3}{x} - \dfrac{1}{3}} \right):\left( {1 + \dfrac{2}{5} + \dfrac{1}{3}} \right) =  - \dfrac{5}{4}\\
 \Leftrightarrow \left( { - 3 - \dfrac{1}{3} - \dfrac{3}{x}} \right):\dfrac{{26}}{{15}} =  - \dfrac{5}{4}\\
 \Leftrightarrow  - \dfrac{{10}}{3} - \dfrac{3}{x} =  - \dfrac{5}{4}.\dfrac{{26}}{{15}}\\
 \Leftrightarrow  - \dfrac{{10}}{3} - \dfrac{3}{x} =  - \dfrac{{13}}{6}\\
 \Leftrightarrow \dfrac{3}{x} =  - \dfrac{{10}}{3} + \dfrac{{13}}{6}\\
 \Leftrightarrow \dfrac{3}{x} = \dfrac{{ - 7}}{6}\\
 \Leftrightarrow x = 3.\dfrac{{ - 6}}{7}\\
 \Leftrightarrow x =  - \dfrac{{18}}{7}\\
Vay\,x =  - \dfrac{{18}}{7}\\
b){\left( {4x - 1} \right)^2} = {\left( {1 - 4x} \right)^4}\\
 \Leftrightarrow {\left( {4x - 1} \right)^2} = {\left( {4x - 1} \right)^4}\\
 \Leftrightarrow {\left( {4x - 1} \right)^4} - {\left( {4x - 1} \right)^2} = 0\\
 \Leftrightarrow {\left( {4x - 1} \right)^2}.\left[ {{{\left( {4x - 1} \right)}^2} - 1} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
4x - 1 = 0\\
{\left( {4x - 1} \right)^2} = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
4x - 1 = 1\\
4x - 1 =  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
4x = 2 \Leftrightarrow x = \dfrac{1}{2}\\
4x = 0 \Leftrightarrow x = 0
\end{array} \right.\\
Vậy\,x \in \left\{ {0;\dfrac{1}{4};\dfrac{1}{2}} \right\}
\end{array}$