ĐKXĐ : \(x\ge1\)
\(\Rightarrow x+3-x+1+2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)
\(\Rightarrow2\sqrt{x^2+2x-3}=2x-2\)
\(\Rightarrow4\left(x^2+2x-3\right)=4x^2-8x+4\)
\(\Rightarrow4x^2+8x-12=4x^2-8x+4\)
\(\Rightarrow16x=16\Rightarrow x=1\)
Vậy x = 1