Question

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

Answer 1

ĐKXĐ: ....

Đặt \(a=\frac{x}{\sqrt{x^2-1}}\) phương trình trở thành: \(x+a=2\sqrt{2}\)

Ta có \(a^2=\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}=1+\frac{1}{x^2-1}\Leftrightarrow a^2-1=\frac{1}{x^2-1}\)

\(\Leftrightarrow\left(a^2-1\right)\left(x^2-1\right)=1\)

\(\Leftrightarrow\left(ax\right)^2-a^2-x^2=0\)

\(\Leftrightarrow\left(ax\right)^2-\left(a+x\right)^2+2ax=0\)

\(\Leftrightarrow\left(ax\right)^2+2ax-8=0\) \(\Rightarrow\left[{}\begin{matrix}ax=2\\ax=-4\end{matrix}\right.\)

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