Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^2} - 6x + 8 = 0\\
\Leftrightarrow \left( {{x^2} - 2x} \right) - \left( {4x - 8} \right) = 0\\
\Leftrightarrow x\left( {x - 2} \right) - 4\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x - 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 4
\end{array} \right.\\
b,\\
2{x^2} + 2x + \dfrac{1}{2} = 0\\
\Leftrightarrow {x^2} + x + \dfrac{1}{4} = 0\\
\Leftrightarrow {x^2} + 2.x.\dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} = 0\\
\Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0\\
\Leftrightarrow x + \dfrac{1}{2} = 0\\
\Leftrightarrow x = - \dfrac{1}{2}\\
c,\\
5x\left( {x - 2015} \right) - x + 2015 = 0\\
\Leftrightarrow 5x\left( {x - 2015} \right) - \left( {x - 2015} \right) = 0\\
\Leftrightarrow \left( {x - 2015} \right)\left( {5x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2015 = 0\\
5x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2015\\
x = \dfrac{1}{5}
\end{array} \right.\\
d,\\
\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) - 8{y^3} + 27 = 0\\
\Leftrightarrow \left( {x + 2y} \right)\left( {{x^2} - x.2y + {{\left( {2y} \right)}^2}} \right) - 8{y^3} + 27 = 0\\
\Leftrightarrow {x^3} + {\left( {2y} \right)^3} - 8{y^3} + 27 = 0\\
\Leftrightarrow {x^3} + 8{y^3} - 8{y^3} + 27 = 0\\
\Leftrightarrow {x^3} + 27 = 0\\
\Leftrightarrow {x^3} = - 27\\
\Leftrightarrow x = - 3\\
e,\\
{x^3} - 4{x^2} + 3x - 12 = 0\\
\Leftrightarrow \left( {{x^3} - 4{x^2}} \right) + \left( {3x - 12} \right) = 0\\
\Leftrightarrow {x^2}\left( {x - 4} \right) + 3.\left( {x - 4} \right) = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {{x^2} + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
{x^2} + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
{x^2} = - 3\,\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow x = 4\\
f,\\
{x^3} + {x^2} - 2x - 8 = 0\\
\Leftrightarrow \left( {{x^3} - 2{x^2}} \right) + \left( {3{x^2} - 6x} \right) + \left( {4x - 8} \right) = 0\\
\Leftrightarrow {x^2}\left( {x - 2} \right) + 3x\left( {x - 2} \right) + 4.\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 3x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
{x^2} + 3x + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
\left( {{x^2} + 3x + \dfrac{9}{4}} \right) + \dfrac{7}{2} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
{\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{7}{2} = 0\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
\Leftrightarrow x = 2
\end{array}\)
