Giải thích các bước giải:
\(\begin{array}{l}
C17.\\
a)\,x\left( {x + 5} \right) - x - 5 = x\left( {x + 5} \right) - \left( {x + 5} \right) = \left( {x + 5} \right)\left( {x - 1} \right)\\
b)\,{x^2} - 2xy - 4 + {y^2} = {x^2} - 2xy + {y^2} - {2^2} = {\left( {x - y} \right)^2} - {2^2} = \left( {x - y + 2} \right)\left( {x - y - 2} \right)\\
C18.\\
a)\,{x^2} - x = 0\\
\Leftrightarrow x\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
b)\,{\left( {x - 2} \right)^2} - 3\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 2 - 3} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x - 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 5
\end{array} \right.
\end{array}\)
