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