Ta có: \(x+2\sqrt{2}.x^2+2x^3=0\)
\(\Leftrightarrow x\left(1+2\sqrt{2}.x+2x^2\right)=0\)
\(\Leftrightarrow x\left[1^2+2.x\sqrt{2}.1+\left(x\sqrt{2}\right)^2\right]=0\)
\(\Leftrightarrow x\left(1+x\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+x\sqrt{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{\sqrt{2}}\end{cases}}\)
Vậy\(x\in\left\{0;\frac{-1}{\sqrt{2}}\right\}\)
\(x+2\sqrt{2}x^2+2x^3=0\)
\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(x\left(2\sqrt{2}x+1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2\sqrt{2}x+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2x\sqrt{2}}\end{cases}}\)
Thôi nhầm chỗ HĐT rồi , bạn làm giống bạn Lê Tài Bảo Châu nhé ! NHầm
Ta có: x+2√2.x2+2x3=0x+22.x2+2x3=0
⇔x(1+2√2.x+2x2)=0⇔x(1+22.x+2x2)=0
⇔x[12+2.x√2.1+(x√2)2]=0⇔x[1^2+2.x\(\sqrt{ }\)2.1+(x2)2]=0
⇔x(1+x√2)2=0⇔x(1+x2)2=0
⇔\orbr{x=01+x√2=0⇔\orbr{x=01+x2=0
⇔\orbr{x=0x=−1√2⇔\orbr{x=0x=−12
Vậyx∈{0;−1√2}