Đáp án:
\(\begin{array}{l}
a.\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = 0
\end{array} \right.\\
b.\left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
c.x = - 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {x - \dfrac{1}{2}} \right)\left( {{x^2} + 1} \right).{x^3} = 0\\
\to \left[ \begin{array}{l}
x - \dfrac{1}{2} = 0\\
x = 0
\end{array} \right.\left( {do:{x^2} + 1 > 0\forall x \in R} \right)\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = 0
\end{array} \right.\\
b.3{x^2} - 2x - 1 = 0\\
\to 3{x^2} - 3x + x - 1 = 0\\
\to 3x\left( {x - 1} \right) + \left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
3x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
c.{x^2} + 2x + 1 = 0\\
\to {\left( {x + 1} \right)^2} = 0\\
\to x + 1 = 0\\
\to x = - 1
\end{array}\)
