Question

Giải phương trình: \(x+\dfrac{2\sqrt{2}x}{\sqrt{1+x^2}}=1\)

Answer 1

\(pt\Leftrightarrow\dfrac{2\sqrt{2}x}{\sqrt{1+x^2}}=1-x\)

\(\Rightarrow\dfrac{8x^2}{1+x^2}=\left(1-x\right)^2\Leftrightarrow x^4-2x^3-6x^2-2x+1=0\)

\(\Leftrightarrow x^2-2x-6-\dfrac{2}{x}+\dfrac{1}{x^2}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=4\\x+\dfrac{1}{x}=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+1=0\\x^2+2x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\left(ktm\right)\\x=2-\sqrt{3}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\) 

Vậy \(x=2-\sqrt{3}\)