Giải thích các bước giải:
\(\begin{array}{l}
b,\\
\left( {x - 2} \right)\left( {2x - 1} \right) = 5\left( {x - 2} \right)\\
\Leftrightarrow \left( {x - 2} \right)\left( {2x - 1} \right) - 5\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right).\left[ {\left( {2x - 1} \right) - 5} \right] = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {2x - 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
2x - 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
c,\\
{\left( {2x - 1} \right)^2} - 4.{\left( {x + 2} \right)^2} = 5\\
\Leftrightarrow {\left( {2x - 1} \right)^2} - {\left[ {2.\left( {x + 2} \right)} \right]^2} = 5\\
\Leftrightarrow {\left( {2x - 1} \right)^2} - {\left( {2x + 4} \right)^2} = 5\\
\Leftrightarrow \left[ {\left( {2x - 1} \right) - \left( {2x + 4} \right)} \right]\left[ {\left( {2x - 1} \right) + \left( {2x + 4} \right)} \right] = 5\\
\Leftrightarrow \left( { - 5} \right).\left( {4x + 3} \right) = 5\\
\Leftrightarrow 4x + 3 = - 1\\
\Leftrightarrow x = - 1\\
d,\\
\frac{{x + 5}}{{x - 5}} - \frac{{x - 5}}{{x + 5}} = \frac{{x\left( {x + 25} \right)}}{{{x^2} - 25}}\,\,\,\,\,\,\,\,\,\left( {x \ne \pm 5} \right)\\
\Leftrightarrow \frac{{{{\left( {x + 5} \right)}^2} - {{\left( {x - 5} \right)}^2}}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \frac{{x\left( {x + 25} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
\Leftrightarrow {\left( {x + 5} \right)^2} - {\left( {x - 5} \right)^2} = x\left( {x + 25} \right)\\
\Leftrightarrow {x^2} + 10x + 25 - {x^2} + 10x - 25 = {x^2} + 25x\\
\Leftrightarrow 20x = {x^2} + 25x\\
\Leftrightarrow {x^2} + 5x = 0\\
\Leftrightarrow x\left( {x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.\\
e,\\
\frac{{x - 5}}{{2015}} + \frac{{x - 4}}{{2016}} = \frac{{x - 3}}{{2017}} + \frac{{x - 2}}{{2018}}\\
\Leftrightarrow \left( {\frac{{x - 5}}{{2015}} - 1} \right) + \left( {\frac{{x - 4}}{{2016}} - 1} \right) = \left( {\frac{{x - 3}}{{2017}} - 1} \right) + \left( {\frac{{x - 2}}{{2018}} - 1} \right)\\
\Leftrightarrow \frac{{x - 5 - 2015}}{{2015}} + \frac{{x - 4 - 2016}}{{2016}} = \frac{{x - 3 - 2017}}{{2017}} + \frac{{x - 2 - 2018}}{{2018}}\\
\Leftrightarrow \frac{{x - 2020}}{{2015}} + \frac{{x - 2020}}{{2016}} = \frac{{x - 2020}}{{2017}} + \frac{{x - 2020}}{{2018}}\\
\Leftrightarrow \left( {x - 2020} \right)\left( {\frac{1}{{2015}} + \frac{1}{{2016}} - \frac{1}{{2017}} - \frac{1}{{2018}}} \right) = 0\\
\Leftrightarrow x - 2020 = 0\\
\Leftrightarrow x = 2020
\end{array}\)
