Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\dfrac{{2x - 1}}{{5 - 3x}} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne \dfrac{5}{3}} \right)\\
\Leftrightarrow 2x - 2 = 2.\left( {5 - 3x} \right)\\
\Leftrightarrow 2x - 2 = 10 - 6x\\
\Leftrightarrow 2x + 6x = 10 + 2\\
\Leftrightarrow 8x = 12\\
\Leftrightarrow x = \dfrac{3}{2}\\
b,\\
\dfrac{{2x + 5}}{{2x}} - \dfrac{x}{{x + 5}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ne 0\\
x \ne - 5
\end{array} \right)\\
\Leftrightarrow \dfrac{{\left( {2x + 5} \right).\left( {x + 5} \right) - 2x.x}}{{2x.\left( {x + 5} \right)}} = 0\\
\Leftrightarrow \dfrac{{\left( {2{x^2} + 15x + 25} \right) - 2{x^2}}}{{2x\left( {x + 5} \right)}} = 0\\
\Leftrightarrow \dfrac{{15x + 25}}{{2x\left( {x + 5} \right)}} = 0\\
\Leftrightarrow 15x + 25 = 0\\
\Leftrightarrow x = - \dfrac{5}{3}\\
c,\\
\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = \dfrac{{16}}{{{x^2} - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne \pm 1} \right)\\
\Leftrightarrow \dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{16}}{{{x^2} - 1}}\\
\Leftrightarrow \dfrac{{\left( {{x^2} + 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)}}{{{x^2} - 1}} = \dfrac{{16}}{{{x^2} - 1}}\\
\Leftrightarrow 4x = 16\\
\Leftrightarrow x = 4\\
d,\\
\dfrac{1}{{3 - x}} - \dfrac{1}{{x + 1}} = \dfrac{x}{{x - 3}} - \dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} - 2x - 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ne 3\\
x \ne - 1
\end{array} \right)\\
\Leftrightarrow - \dfrac{1}{{x - 3}} - \dfrac{1}{{x + 1}} = \dfrac{x}{{x - 3}} - \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
\Leftrightarrow \dfrac{{ - \left( {x + 1} \right) - \left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 1} \right)}} = \dfrac{{x.\left( {x + 1} \right) - {{\left( {x - 1} \right)}^2}}}{{\left( {x + 1} \right).\left( {x - 3} \right)}}\\
\Leftrightarrow - x - 1 - x + 3 = {x^2} + x - \left( {{x^2} - 2x + 1} \right)\\
\Leftrightarrow - 2x + 2 = 3x - 1\\
\Leftrightarrow 5x = 3\\
\Leftrightarrow x = \dfrac{3}{5}\\
2,\\
\dfrac{8}{{x - 8}} + \dfrac{{11}}{{x - 11}} = \dfrac{9}{{x - 9}} + \dfrac{{10}}{{x - 10}}\,\,\,\,\,\,\,\,\,\,\left( {x \ne 8;\,x \ne 9;\,x \ne 10;\,x \ne 11} \right)\\
\Leftrightarrow \left( {\dfrac{8}{{x - 8}} + 1} \right) + \left( {\dfrac{{11}}{{x - 11}} + 1} \right) = \left( {\dfrac{9}{{x - 9}} + 1} \right) + \left( {\dfrac{{10}}{{x - 10}} + 1} \right)\\
\Leftrightarrow \dfrac{{8 + \left( {x - 8} \right)}}{{x - 8}} + \dfrac{{11 + \left( {x - 11} \right)}}{{x - 11}} = \dfrac{{9 + \left( {x - 9} \right)}}{{x - 9}} + \dfrac{{10 + \left( {x - 10} \right)}}{{x - 10}}\\
\Leftrightarrow \dfrac{x}{{x - 8}} + \dfrac{x}{{x - 11}} = \dfrac{x}{{x - 9}} + \dfrac{x}{{x - 10}}\\
\Leftrightarrow x.\left( {\dfrac{1}{{x - 8}} + \dfrac{1}{{x - 11}} - \dfrac{1}{{x - 9}} - \dfrac{1}{{x - 10}}} \right) = 0\\
\Leftrightarrow x = 0
\end{array}\)
