Question

Giải phương trình

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

Answer 1

Lời giải:

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

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

\(\Leftrightarrow x^2-7+3(x+2)=\sqrt{x^2+2}(x+2)\)

\(\Leftrightarrow x^2-7=(x+2)(\sqrt{x^2+2}-3)\)

\(\Leftrightarrow x^2-7=\frac{(x+2)(x^2-7)}{\sqrt{x^2+2}+3}\)

\(\Leftrightarrow (x^2-7)\left(1-\frac{x+2}{\sqrt{x^2+2}+3}\right)=0\)

TH1: \(x^2-7=0\Leftrightarrow x=\pm \sqrt{7}\) (thỏa mãn)

TH2: \(1-\frac{x+2}{\sqrt{x^2+2}+3}=0\Leftrightarrow x+2=\sqrt{x^2+2}+3\)

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

\(\Rightarrow x^2-2x+1=x^2+2\Leftrightarrow x=-\frac{1}{2}\) (vô lý vì \(x\geq 1\) )

Vậy \(x=\pm \sqrt{7}\)