\(x^3-2x=-x^2+2\)
\(\Leftrightarrow x^3+x^2-2x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Ta có: \(x^3-2x=-x^2+2\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(2x+2\right)=0\)
\(\Leftrightarrow x^2.\left(x+1\right)-2.\left(x+1\right)=0\)
5\(\Leftrightarrow\left(x+1\right).\left(x^2-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{2}\end{cases}}\)
Vậy \(S=\left\{-\sqrt{2};-1;\sqrt{2}\right\}\)
\(x^3-2x=-x^2+2\)
\(x^3+x^2-2x-2=0\)
\(x^2\left(x+1\right)-2\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2-2\right)=0\)
\(Th1:x+1=0\Leftrightarrow x=-1\)
\(Th2:x^2-2=0\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)
\(x^3-2x=-x^2+2\)
\(=>x^3-2x+x^2-2=0\)
\(=>x^2.\left(x-1\right)-2.\left(x-1\right)=0\)
\(=>\left(x^2-2\right)\left(x-1\right)=0\)
\(=>\orbr{\begin{cases}x-1=0\\x^2-2=0\end{cases}=>\orbr{\begin{cases}x=1\\x=\pm\sqrt{2}\end{cases}}}\)
