ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
- Với \(x=-\dfrac{1}{2}\) là nghiệm
- Với \(x>-\dfrac{1}{2}\)
\(\Rightarrow x+1-\sqrt{x^2+8x+4}+2\sqrt{2x+1}=0\)
\(\Leftrightarrow\dfrac{-3\left(2x+1\right)}{x+1+\sqrt{x^2+8x+4}}+\dfrac{2\left(2x+1\right)}{\sqrt{2x+1}}=0\)
\(\Leftrightarrow\dfrac{-3}{x+1+\sqrt{x^2+8x+4}}+\dfrac{2}{\sqrt{2x+1}}=0\)
\(\Leftrightarrow2x+2+2\sqrt{x^2+8x+4}-3\sqrt{2x+1}=0\)
\(\Leftrightarrow\left(2x+1-2\sqrt{2x+1}+1\right)+\left(2\sqrt{x^2+4\left(2x+1\right)}-\sqrt{2x+1}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-1\right)^2+\dfrac{4x^2+15\left(2x+1\right)}{2\sqrt{x^2+4\left(2x+1\right)}+\sqrt{2x+1}}=0\) (vô nghiệm do vế trái luôn dương)
Vậy \(x=-\dfrac{1}{2}\) là nghiệm duy nhất
