Đáp án:$x \in \left\{ {1; - 0,17; - 2,82} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
P\left( x \right) = 2{x^3} + 4{x^2} - 5x - 1\\
= 2{x^3} - 2{x^2} + 6{x^2} - 6x + x - 1\\
= 2{x^2}\left( {x - 1} \right) + 6x\left( {x - 1} \right) + \left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {2{x^2} + 6x + 1} \right)\\
= 2.\left( {x - 1} \right)\left( {{x^2} + 3x + \frac{1}{2}} \right)\\
Khi:P\left( x \right) = 0\\
\Rightarrow 2.\left( {x - 1} \right).\left( {{x^2} + 3x + \frac{1}{2}} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 0\\
{x^2} + 3x + \frac{1}{2} = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - 0,17\\
x = - 2,82
\end{array} \right.
\end{array}$
Vậy $x \in \left\{ {1; - 0,17; - 2,82} \right\}$
