\(DK:x>-\frac{1}{2}\)
Dat \(\sqrt{x^2+2x+3}=t\ge\sqrt{2}\)
PT tro thanh
\(t^2-\left(2x+1\right)t+4x-2=0\)
Ta co:
\(\Delta_t=\left(2x-3\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}t_1=2x-1\\t_2=2\\t_3=x+\frac{1}{2}\end{cases}}\)
Sau do the vao roi giai la xong :D
pt <=> \(\left(x^2+2x+3\right)-\left(2x+1\right)\sqrt{x^2+2x+3}+4x-2=0\)
đặt t=\(\sqrt{x^2+2x+3}\left(t\ge3\right)\), ta được \(r^2-\left(2x+1\right)t+4x-2=0\)
ta có: \(\Delta=\left(2x-3\right)^2\)=> pt có 2 nghiệm t=2x-1; t=2
với t=2x-1 ta có: \(\sqrt{x^2+2x+3}=2x-1\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\3x^2-6x-2=0\end{cases}\Leftrightarrow x=\frac{3+\sqrt{5}}{3}}\)
với t=2 ta có: \(\sqrt{x^2+2x+3}=2\Leftrightarrow x^2+2x-1=0\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)
Vậy....
