\(\sqrt{x^2+2x}+\sqrt{2x-1}=\sqrt{3x^2+4x+1}\)(ĐK:\(x>\frac{1}{2}\))
\(\Leftrightarrow x^2+2x+2x-1+2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}=3x^2+4x+1\)(BP 2 vế)
\(\Leftrightarrow2\sqrt{2x^3-x^2+4x^2-2x}=2x^2+2\)
\(\Leftrightarrow\sqrt{2x^3+2x+3x^2+3-4x-3}=x^2+1\)
Đặt \(x^2+1=t\)
pt\(\Leftrightarrow\sqrt{2xt+3t-\left(4x+3\right)}=t\)
\(\Leftrightarrow2xt+3t-4x-3=t^2\)
\(\Leftrightarrow t^2-t\left(2x+3\right)+4x+3=0\)
\(\Delta=\left(2x+3\right)^2-4.\left(4x+3\right)=4x^2+12x+9-16x-12=4x^2-4x-3\)
\(\hept{\begin{cases}t_1=\frac{2x+3-\sqrt{4x^2-4x-3}}{2}\\t_2=\frac{2x+3+\sqrt{4x^2-4x-3}}{2}\end{cases}}\)
TH1:\(t=\frac{2x+3-\sqrt{4x^2-4x-3}}{2}\)
\(\Rightarrow2x^2+2=2x+3-\sqrt{4x^2-4x-3}\)
\(\Leftrightarrow2x^2+2=2x+3-\sqrt{4x^2+4x-8x-3}\)
\(\Leftrightarrow2t=2x+3-\sqrt{4t-8x-3}\)
Giải ra rồi thay TH2
