\(x+2\sqrt{2}x^2+2x^3=0\) (1)
\(\Leftrightarrow x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(\Leftrightarrow x\left(1+\sqrt{2}x\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(1+\sqrt{2}x\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{\sqrt{2}}{2};0\right\}\)
\(x+2\sqrt{2}x^2+2x^3=0\Leftrightarrow x\left(2x^2+2.\sqrt{2}x+1\right)=0\Leftrightarrow x\left(\sqrt{2}x+1\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{2}x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{2}x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{\sqrt{2}}\end{matrix}\right.\)
\(x+2\sqrt{2}x^2+2x^3=0\)
\(\Leftrightarrow x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(\Leftrightarrow x\left(1+\sqrt{2}x\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
Vậy x1 = 0 ; x2 = \(-\dfrac{\sqrt{2}}{2}\)
