Giải hệ phương trình \(\left\{\begin{matrix} (1-y)\sqrt{x-y}+x=2+(x-y-1)\sqrt{y}

Điều kiện: \(\left\{\begin{matrix} y\geq 0\\ x\geq y \end{matrix}\right.\Leftrightarrow x\geq y\geq 0 \ \ (*)\)
\((1)\Leftrightarrow (1-y)\sqrt{x-y}+y-1+x-y-1-(x-y-1)\sqrt{y}=0\)
\(\Leftrightarrow (1-y)(\sqrt{x-y}-1)+(x-y-1)(1-\sqrt{x})=0\)
\(\Leftrightarrow (1-y)(x-y-1)\underbrace{ \left ( \frac{1}{\sqrt{x-y}+1}+\frac{1}{1+\sqrt{y}} \right )}_{>0}=0\Leftrightarrow \bigg \lbrack\begin{matrix} y=1\\ y=x-1 \end{matrix}\)
TH1: \(y=1\) mà \(x\geq y\) nên \(x\geq 1\)
\((2)\Leftrightarrow 5x+7=(x+5)\sqrt{x^2+3}\)
\(\Leftrightarrow 25x^2+70x+49=(x^2+10x+25)(x^2+3)\)
\(\Leftrightarrow x^4+10x^3+3x^2-40x+26=0\)
\(\Leftrightarrow (x-1)^2(x^2+12x+26)=0\Leftrightarrow x=1\) ( do \(x\geq 1\))
TH2: y = x - 1
\(\small (2)\Leftrightarrow 2x^2+5x-1=7\sqrt{(x-1)(x^2+x+1)} \ (3)\)
 Đặt \(\small \sqrt{x-1}=A; \sqrt{x^2+x+1}=b\ (A\geq 0;B\geq 0)\)
Ta có \(\small 2B^2+3A^2=2x^2+5x-1\)
\(\small (3)\Leftrightarrow 2B^2+3A^2=7AB\Leftrightarrow 2B^2-7AB+3A^2=0\Leftrightarrow \bigg \lbrack\begin{matrix} B=3A\\ A=2B \end{matrix}\)
\(\small \Leftrightarrow \bigg \lbrack\begin{matrix} \sqrt{x^2+x+1}=3\sqrt{x-1}\\ \sqrt{x-1}=2\sqrt{x^2+x+1} \end{matrix}\Leftrightarrow \bigg \lbrack\begin{matrix} x^2-8x=10=0\\ 4x^2+3x+5=0 \end{matrix}\)
\(\small \Leftrightarrow \bigg \lbrack\begin{matrix} x=4+\sqrt{6}\Rightarrow y=3+\sqrt{6}\\ x=4-\sqrt{6}\Rightarrow y=3-\sqrt{6} \end{matrix}\)
Vậy hệ có nghiệm là \(\small (1;1),(4+\sqrt{6};3+\sqrt{6}),(4-\sqrt{6};3-\sqrt{6})\)