Question

giải phương trình : \(x^3-3x^2+2\sqrt{2}x+2-2\sqrt{2}=0\)

Answer 1

\(x^3-3x^2+2\sqrt{2}x+2-2\sqrt{2}=0\)

\(x^3-3x^2+3x-1=3x-2\sqrt{2}x-3+2\sqrt{2}\)

\(\left(x-1\right)^3=\left(3-2\sqrt{2}\right)x-\left(3-2\sqrt{2}\right)\)

\(\left(x-1\right)^3=\left(3-2\sqrt{2}\right)\left(x-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0;x=1\\\left(x-1\right)^2=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1-\left(\sqrt{2}-1\right)=2-\sqrt{2}\\x=1+\left(\sqrt{2}-1\right)=\sqrt{2}\end{matrix}\right.\)