Đáp án:
$\begin{array}{l}
a)2{x^3} + 10x = 0\\
\Rightarrow 2x\left( {{x^2} + 5} \right) = 0\\
\Rightarrow x = 0\left( {do:{x^2} + 5 > 0} \right)\\
\text{Vậy}\,x = 0\\
b)2{x^2}\left( {x - 3} \right) - 2x + 6 = 0\\
\Rightarrow 2{x^2}\left( {x - 3} \right) - 2\left( {x - 3} \right) = 0\\
\Rightarrow \left( {x - 3} \right)\left( {2{x^2} - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 3 = 0\\
2{x^2} - 2 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
{x^2} = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1\\
x = 1
\end{array} \right.\\
\text{Vậy}\,x = - 1;x = 1;x = 3\\
c){\left( {x - 2} \right)^2} = {\left( {x + 1} \right)^2}\\
\Rightarrow {\left( {x - 2} \right)^2} - {\left( {x + 1} \right)^2} = 0\\
\Rightarrow \left( {x - 2 - x - 1} \right)\left( {x - 2 + x + 1} \right) = 0\\
\Rightarrow - 3.\left( {2x - 1} \right) = 0\\
\Rightarrow 2x - 1 = 0\\
\Rightarrow x = \dfrac{1}{2}\\
\text{Vậy}\,x = \dfrac{1}{2}\\
d)3{x^2} - 7x + 4 = 0\\
\Rightarrow 3{x^2} - 3x - 4x + 4 = 0\\
\Rightarrow 3x\left( {x - 1} \right) - 4\left( {x - 1} \right) = 0\\
\Rightarrow \left( {x - 1} \right)\left( {3x - 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 0\\
3x - 4 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{4}{3}
\end{array} \right.\\
\text{Vậy}\,x = 1;x = \dfrac{4}{3}
\end{array}$
