Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{x - 1}}{{x - 2}} = \dfrac{4}{5}\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne 2} \right)\\
\Leftrightarrow 5.\left( {x - 1} \right) = 4.\left( {x - 2} \right)\\
\Leftrightarrow 5x - 5 = 4x - 8\\
\Leftrightarrow 5x - 4x = - 8 + 5\\
\Leftrightarrow x = - 3\\
b,\\
\dfrac{{x + 7}}{{ - 20}} = \dfrac{{ - 5}}{{x + 7}}\,\,\,\,\,\,\,\,\,\left( {x \ne - 7} \right)\\
\Leftrightarrow \left( {x + 7} \right)\left( {x + 7} \right) = \left( { - 20} \right).\left( { - 5} \right)\\
\Leftrightarrow {\left( {x + 7} \right)^2} = 100\\
\Leftrightarrow \left[ \begin{array}{l}
x + 7 = 10\\
x + 7 = - 10
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 17
\end{array} \right.\\
c,\\
\dfrac{x}{8} = \dfrac{x}{{{x^3}}}\,\,\,\,\,\,\left( {x \ne 0} \right)\\
\Leftrightarrow \dfrac{x}{8} = \dfrac{1}{{{x^2}}}\\
\Leftrightarrow x.{x^2} = 8\\
\Leftrightarrow {x^3} = {2^3}\\
\Leftrightarrow x = 2\\
d,\\
{2^{2x + 1}} + {4^{x + 3}} = 264\\
\Leftrightarrow {2^{2x}}{.2^1} + {4^x}{.4^3} = 264\\
\Leftrightarrow {4^x}.2 + {4^x}.64 = 264\\
\Leftrightarrow {66.4^x} = 264\\
\Leftrightarrow {4^x} = 264:66\\
\Leftrightarrow {4^x} = 4\\
\Leftrightarrow x = 1
\end{array}\)
