\(x+2\sqrt{2x^2}+2x^3=0\)
\(\Leftrightarrow x+2x\sqrt{2}+2x^3=0\)
\(\Leftrightarrow x\left(1+2\sqrt{2}+2x^2\right)=0\)
\(\Leftrightarrow x=0\) ( Vì \(1+2\sqrt{2}+2x^2>0\) )
Tìm x biết :
\(x+2\sqrt{2}x^2+2x^3=0\)
\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(x\left(1+\sqrt{2}x\right)^2=0\)
TH1 : x=0
TH2 : \(\left(1+\sqrt{2}x\right)^2=0\)
\(1+\sqrt{2}x=0\)
\(x=\frac{-1}{\sqrt{2}}\)
Bạn sửa lại đề nha, sai đó
\(x+2\sqrt{2}x^2+2x^3=0\)
\(\Rightarrow x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(\Rightarrow\left(1+\sqrt{2}x\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(1+\sqrt{2}x\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
Vậy x = 0 hoặc x = \(-\dfrac{\sqrt{2}}{2}\)
