Question

giải phương trình:

\(\sqrt{2x^2+1}+2\sqrt{x+1}=\sqrt{2x^2+x+9}\)

Answer 1

\(DK:x\ge-1\)

\(\Leftrightarrow\left(\sqrt{2x^2+1}-1\right)+\left(2\sqrt{x+1}-2\right)+\left(3-\sqrt{2x^2+x+9}\right)=0\)\(\Leftrightarrow\frac{2x^2}{\sqrt{2x^2+1}+1}+\frac{2x}{\sqrt{x+1}+1}-\frac{x\left(2x+1\right)}{3+\sqrt{2x^2+x+9}}=0\)

\(\Leftrightarrow x\left(\frac{2x}{\sqrt{2x^2+1}}+\frac{2}{\sqrt{x+1}+1}-\frac{2x+1}{3\sqrt{2x^2+x+9}}\right)=0\)

Cai ngoac to thi >0 nhung khong biet danh gia sao :)

\(\Rightarrow x=0\left(n\right)\)

Vay nghiem cua PT la \(x=0\)