Đáp án:
$\begin{array}{l}
1)\dfrac{{x - 1}}{3} - \dfrac{{5x + 2}}{4} = \dfrac{x}{5}\\
\Rightarrow \dfrac{{20\left( {x - 1} \right) - 15\left( {5x + 2} \right)}}{{60}} = \dfrac{{12x}}{{60}}\\
\Rightarrow 20x - 20 - 75x - 30 = 12x\\
\Rightarrow 12x + 75x - 20x = - 50\\
\Rightarrow 67x = - 50\\
\Rightarrow x = \dfrac{{ - 50}}{{67}}\\
2)Dkxd:x \ne \dfrac{1}{4};x \ne - \dfrac{1}{4}\\
\dfrac{3}{{1 - 4x}} = \dfrac{2}{{4x + 1}} - \dfrac{{8 - 6x}}{{16{x^2} - 1}}\\
\Rightarrow \dfrac{{2\left( {4x - 1} \right) - 8 + 6x + 3\left( {4x + 1} \right)}}{{\left( {4x - 1} \right)\left( {4x + 1} \right)}} = 0\\
\Rightarrow 8x - 2 - 8 + 6x + 12x + 3 = 0\\
\Rightarrow 26x - 7 = 0\\
\Rightarrow x = \dfrac{7}{{26}}\left( {tmdk} \right)\\
3)\dfrac{1}{{x - 1}} - \dfrac{{3{x^2}}}{{x - 1}} = \dfrac{{2x}}{{x - 1}}\left( {dkxd:x \ne 1} \right)\\
\Rightarrow 1 - 3{x^2} = 2x\\
\Rightarrow 3{x^2} + 2x - 1 = 0\\
\Rightarrow \left( {3x - 1} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{3}\left( {tmdk} \right)\\
x = - 1\left( {tmdk} \right)
\end{array} \right.\\
5)\dfrac{1}{{1 - 2x}} \le 0\\
Do:1 > 0\\
\Rightarrow 1 - 2x < 0\\
\Rightarrow 2x > 1\\
\Rightarrow x > \dfrac{1}{2}
\end{array}$
Câu 4 khó nhìn đề bài quá.
