Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
5x\left( {x - y} \right) + 4.\left( {y - x} \right) = 5x.\left( {x - y} \right) - 4.\left( {x - y} \right)\\
= \left( {x - y} \right).\left( {5x - 4} \right)\\
b,\\
{x^2} - 5x + 5xy - 25y = \left( {{x^2} - 5x} \right) + \left( {5xy - 25y} \right)\\
= x.\left( {x - 5} \right) + 5y.\left( {x - 5} \right) = \left( {x - 5} \right)\left( {x + 5y} \right)\\
2,\\
a,\\
{\left( {x + 1} \right)^2} - x\left( {x - 2} \right) = 3\\
\Leftrightarrow \left( {{x^2} + 2x + 1} \right) - \left( {{x^2} - 2x} \right) = 3\\
\Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x = 3\\
\Leftrightarrow 4x + 1 = 3\\
\Leftrightarrow 4x = - 2\\
\Leftrightarrow x = - \frac{1}{2}\\
b,\\
{\left( {x + 2} \right)^2} + \left( {3 - x} \right)\left( {2 + x} \right) = 1\\
\Leftrightarrow \left( {x + 2} \right).\left[ {\left( {x + 2} \right) + \left( {3 - x} \right)} \right] = 1\\
\Leftrightarrow \left( {x + 2} \right).5 = 1\\
\Leftrightarrow x + 2 = \frac{1}{5}\\
\Leftrightarrow x = - \frac{9}{5}\\
c,\\
{\left( {2x - 3} \right)^2} = {\left( {x + 5} \right)^2}\\
\Leftrightarrow {\left( {2x - 3} \right)^2} - {\left( {x + 5} \right)^2} = 0\\
\Leftrightarrow \left[ {\left( {2x - 3} \right) - \left( {x + 5} \right)} \right].\left[ {\left( {2x - 3} \right) + \left( {x + 5} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 8} \right)\left( {3x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 8 = 0\\
3x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = - \frac{2}{3}
\end{array} \right.\\
d,\\
{x^2}\left( {x - 1} \right) - 4{x^2} + 8x - 4 = 0\\
\Leftrightarrow {x^2}\left( {x - 1} \right) - 4.\left( {{x^2} - 2x + 1} \right) = 0\\
\Leftrightarrow {x^2}\left( {x - 1} \right) - 4.{\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow \left( {x - 1} \right).\left[ {{x^2} - 4.\left( {x - 1} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right).\left( {{x^2} - 4x + 4} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right){\left( {x - 2} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)
