Question

Giải phương trình:

\(x+\dfrac{2x\sqrt{6}}{\sqrt{x^2}+1}=1\)

Answer 1

Ta có \(x+\dfrac{2x\sqrt{6}}{\sqrt{x^2}+1}=1\Rightarrow x+\dfrac{2x\sqrt{6}}{x+1}=1\Rightarrow\dfrac{x^2+x+2x\sqrt{6}}{x+1}=1\Rightarrow x^2+2x\sqrt{6}=1\Rightarrow x^2+2x\sqrt{6}+6=7\Rightarrow\left(x+\sqrt{6}\right)^2=7\Rightarrow x+\sqrt{6}=\pm\sqrt{7}\Rightarrow x\in\left\{\sqrt{7}-\sqrt{6},-\sqrt{7}-\sqrt{6}\right\}\)Vậy \(x\in\left\{\sqrt{7}-\sqrt{6},-\sqrt{7}-\sqrt{6}\right\}\)