ĐKXĐ: \(x\ge0;x+y-4\ge0\)
\(PT_{\left(1\right)}\Leftrightarrow\left(\sqrt{x^2+3}-2\right)+\left(\sqrt{x}-1\right)+\left(\sqrt{x+3}-2\right)=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+3}+2}+\frac{x-1}{\sqrt{x}+1}+\frac{x-1}{\sqrt{x+3}+2}=0\)
\(\Leftrightarrow\left(x-1\right)\left[\frac{\left(x+1\right)}{\sqrt{x^2+3}+2}+\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x+3}+2}\right]=0\)
Cái ngoặc to vô nghiệm. Vậy x = 1.
Thay xuống PT (2) \(\Leftrightarrow3+\sqrt{y-3}=5\left(Đ\text{K:}y\ge3\right)\Leftrightarrow\sqrt{y-3}=2\Leftrightarrow y=7\)
Vậy x = 1; y = 7
P/s: Em ko chắc.
