Giải thích các bước giải:
\(\begin{array}{l}
a,\\
{\left( {x - 3} \right)^2} - 4{\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow {\left( {x - 3} \right)^2} = {2^2}.{\left( {x + 1} \right)^2}\\
\Leftrightarrow {\left( {x - 3} \right)^2} = {\left( {2x + 2} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 2x + 2\\
x - 3 = - 2x - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 5\\
x = \frac{1}{3}
\end{array} \right.\\
b,\\
5x\left( {x - 1} \right) = 10\left( {1 - x} \right)\\
\Leftrightarrow 5x\left( {x - 1} \right) - 10\left( {1 - x} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {5x + 10} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
5x + 10 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\\
c,\\
\frac{x}{8} - \frac{{x + 3}}{6} = \frac{1}{4}\\
\Leftrightarrow \frac{x}{8} - \frac{x}{6} - \frac{3}{6} = \frac{1}{4}\\
\Leftrightarrow x\left( {\frac{1}{8} - \frac{1}{6}} \right) = \frac{1}{4} + \frac{1}{2}\\
\Leftrightarrow x.\left( { - \frac{1}{{24}}} \right) = \frac{3}{4}\\
\Leftrightarrow x = - 18\\
d,\\
{x^2} - 2x = - 4\\
\Leftrightarrow {x^2} - 2x + 1 = - 3\\
\Leftrightarrow {\left( {x - 1} \right)^2} = - 3\,\,\,\,\left( {VN} \right)\\
e,\\
{x^3} - {x^2} - x + 1 = 0\\
\Leftrightarrow \left( {{x^3} - {x^2}} \right) - \left( {x - 1} \right) = 0\\
\Leftrightarrow {x^2}\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 1} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 1 = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\\
g,\\
{x^2} - 4x + 3 = 0\\
\Leftrightarrow \left( {{x^2} - 3x} \right) - \left( {x - 3} \right) = 0\\
\Leftrightarrow x\left( {x - 3} \right) - \left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.
\end{array}\)
