Tìm x: a/ 2$x^{2}$ + 6x = 0 b/ $x^{5}$ – 16x = 0 c/ 4$x^{4}$ – 9$x^{2}$ = 0 d/ 1/2$x^{2}$ + 3x = 0

Đáp án:

$\begin{array}{l}
a)2{x^2} + 6x = 0\\
 \Leftrightarrow 2x\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 3
\end{array} \right.\\
Vậy\,x =  - 3;x = 0\\
b){x^5} - 16x = 0\\
 \Leftrightarrow x\left( {{x^4} - 16} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^4} = 16
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x =  - 2
\end{array} \right.\\
Vậy\,x =  - 2;x = 0;x = 2\\
c)4{x^4} - 9{x^2} = 0\\
 \Leftrightarrow {x^2}\left( {4{x^2} - 9} \right) = 0\\
 \Leftrightarrow {x^2}\left( {2x - 3} \right)\left( {2x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
2x - 3 = 0\\
2x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{3}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
Vậy\,x =  - \dfrac{3}{2};x = 0;x = \dfrac{3}{2}\\
d)\dfrac{1}{2}{x^2} + 3x = 0\\
 \Leftrightarrow x\left( {\dfrac{1}{2}x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
\dfrac{1}{2}x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 6
\end{array} \right.\\
Vậy\,x =  - 6;x = 0
\end{array}$