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