Đáp án:
$\begin{array}{l}
a)\left( {x - \dfrac{1}{2}} \right).\left( {{x^2} + 1} \right).{x^3} = 0\\
\Rightarrow \left[ \begin{array}{l}
x - \dfrac{1}{2} = 0\\
{x^2} + 1 = 0\\
{x^3} = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
{x^2} = - 1\left( {vô\,nghiệm} \right)\\
x = 0
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2}\,hoặc\,x = 0\\
b)3{x^2} - 2x - 1 = 0\\
\Rightarrow 3{x^2} - 3x + x - 1 = 0\\
\Rightarrow 3x.\left( {x - 1} \right) + \left( {x - 1} \right) = 0\\
\Rightarrow \left( {x - 1} \right).\left( {3x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 0\\
3x + 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x = 1\,hoặc\,x = - \dfrac{1}{3}\\
c){x^2} + 2x + 1 = 0\\
\Rightarrow {x^2} + x + x + 1 = 0\\
\Rightarrow x\left( {x + 1} \right) + \left( {x + 1} \right) = 0\\
\Rightarrow \left( {x + 1} \right).\left( {x + 1} \right) = 0\\
\Rightarrow x + 1 = 0\\
\Rightarrow x = - 1
\end{array}$
Vậy x=-1
