ĐKXĐ: ...
\(\Leftrightarrow3x^2+3x+1=\left(3x+1\right)\sqrt{x^2+2x-1}\)
\(\Leftrightarrow3\left(x^2+2x-1\right)-\left(3x+1\right)\sqrt{x^2+2x-1}-3x-4=0\)
Đặt \(\sqrt{x^2+2x-1}=t\ge0\)
\(3t^2-\left(3x+1\right)t-3x-4=0\)
\(a-b+c=3+3x+1-3x-4=0\)
\(\Rightarrow\) Phương trình luôn có 2 nghiệm: \(\left[{}\begin{matrix}a=-1< 0\left(l\right)\\a=\frac{3x+4}{3}\end{matrix}\right.\)
\(\Rightarrow\frac{3x+4}{3}=\sqrt{x^2+2x-1}\) (với \(x\ge-\frac{4}{3}\))
\(\Leftrightarrow\left(3x+4\right)^2=9\left(x^2+2x-1\right)\)
\(\Leftrightarrow...\)
