Đáp án:
\[\left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^4} + 2{x^3} + 5{x^2} + 4x - 12 = 0\\
\Leftrightarrow \left( {{x^4} - {x^3}} \right) + \left( {3{x^3} - 3{x^2}} \right) + \left( {8{x^2} - 8x} \right) + \left( {12x - 12} \right) = 0\\
\Leftrightarrow {x^3}\left( {x - 1} \right) + 3{x^2}\left( {x - 1} \right) + 8x\left( {x - 1} \right) + 12\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 3{x^2} + 8x + 12} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right).\left[ {\left( {{x^3} + 2{x^2}} \right) + \left( {{x^2} + 2x} \right) + \left( {6x + 12} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right).\left[ {{x^2}\left( {x + 2} \right) + x\left( {x + 2} \right) + 6\left( {x + 2} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} + x + 6} \right) = 0\\
{x^2} + x + 6 = {\left( {x + \frac{1}{2}} \right)^2} + 5\frac{3}{4} > 0,\,\,\,\forall x\\
\Rightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.
\end{array}\)
