\(5x^2-3=0\Leftrightarrow x^2=\dfrac{3}{5}\Leftrightarrow x=\pm\sqrt{\dfrac{3}{5}}=\pm\dfrac{\sqrt{15}}{5}\)
\(4x^3+x=0\Leftrightarrow x\left(4x^2+1\right)=0\Leftrightarrow x=0;4x^2+1>0\)
\(5x^2-3=0\\ \Leftrightarrow5x^2=3\\ \Leftrightarrow x^2=\dfrac{3}{5}\\\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{3}{5}}\\x=-\sqrt{\dfrac{3}{5}}\end{matrix}\right. \)
vậy \(x=\sqrt{\dfrac{3}{5}}\) ;\(x=-\sqrt{\dfrac{3}{5}}\)
\(4x^3+x=0\\ \Leftrightarrow x\left(4x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{-1}{4}\left(vl\right)\end{matrix}\right.\)
vậy x=0
\(5x^2-3=0 \)
<=> 5x2 =3
<=> x2= \(\dfrac{3}{5}\)
<=>\(x=\sqrt{\dfrac{3}{5}}\)hay \(x=-\sqrt{\dfrac{3}{5}}\)
Vậy S={\(\sqrt{\dfrac{3}{5}}\);\(-\sqrt{\dfrac{3}{5}} \)}
4x3+x=0
<=> x(4x2+1)=0
<=>x=0 hay 4x2+1=0
<=> x=0 hay 4x2=-1(vô lý)
Vậy S={0}
a,\(5x^2-3=0\)
\(\Leftrightarrow5x^2=3\)
\(\Leftrightarrow x^2=\dfrac{3}{5}\)
\(\Leftrightarrow x=\pm\sqrt{\dfrac{3}{5}}\) \(=\pm\dfrac{\sqrt{15}}{5}\)
Vậy tập nghiệm của phương trình là S=\(\left\{\pm\dfrac{\sqrt{15}}{5}\right\}\)
b,\(4x^3+x=0\)
\(\Leftrightarrow x\left(4x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{-1}{4}\left(KTM\right)\end{matrix}\right.\)
Vậy tập nghệm của phương trình là S=0
a) Ta có: \(5x^2-3=0\)
\(\Leftrightarrow5x^2=3\)
\(\Leftrightarrow x^2=\dfrac{3}{5}\)
hay \(x\in\left\{\dfrac{\sqrt{15}}{5};-\dfrac{\sqrt{15}}{5}\right\}\)
b) Ta có: \(4x^3+x=0\)
\(\Leftrightarrow x\left(4x^2+1\right)=0\)
\(\Leftrightarrow x=0\)
