\[\begin{array}{l}
\begin{cases} \left(x+2\right)^2-x\left(y+1\right)+y=8\left(1\right)\\ 4x^2-24x+35=5\left(\sqrt{3y-14}+\sqrt{y-1}\right)\left(2\right) \end{cases}\\
ĐK:\,y\ge \dfrac{14}{3}\\
\left(1\right)\leftrightarrow \left[\left(x+2\right)^2-9\right]+\left(y-xy-x+1\right)=0\\
\leftrightarrow \left(x-1\right)\left(x+5\right)-\left(x-1\right)\left(y+1\right)=0\\
\leftrightarrow \left(x-1\right)\left(x-y+4\right)=0\\
\to \left[ \begin{array}{l}x=1\\y=x+4\end{array} \right.\\
Voi\,x=1\,thay\,vao\,PT(2):\\
\left(2\right)\leftrightarrow 4-24+35=5\left(\sqrt{3y-14}+\sqrt{y-1}\right)\\ \leftrightarrow \sqrt{3y-14}=3-\sqrt{y-1}\\ \leftrightarrow 3y-14=y+8-6\sqrt{y-1}\\ \leftrightarrow 11-y=3\sqrt{y-1}\\
\leftrightarrow y^2-22y+121=9y-9\\
\leftrightarrow y^2-31y+130=0\\
\leftrightarrow \left(y-26\right)\left(y-5\right)=0\\
\to y=26;y=5\\
Voi\,y=x+4\,thay\,vao\,PT(2):\\
\leftrightarrow \left(5-5\sqrt{3x-2}\right)+\left(10-5\sqrt{x+3}\right)+4x^2-24x+20=0\\
\leftrightarrow 4\left(x-5\right)\left(x-1\right)-5\left[\dfrac{3\left(x-1\right)}{1+\sqrt{3x-2}}+\dfrac{x-1}{2+\sqrt{x+3}}\right]=0\\
\to \left[\begin{array}{l}x=1\\4\left(x-5\right)-5\left(\dfrac{3}{1+\sqrt{3x-2}}+\dfrac{1}{2+\sqrt{x+3}}\right)=0\left(3\right)\end{array}\right.\\
\left(3\right)\leftrightarrow 4x-20-\dfrac{15}{1+\sqrt{3x-2}}-\dfrac{5}{2+\sqrt{x+3}}=0\\ \leftrightarrow \left(3-\dfrac{15}{1+\sqrt{3x-2}}\right)+\left(1-\dfrac{5}{2+\sqrt{x+3}}\right)+4x-24=0\\
\leftrightarrow \dfrac{3\left(\sqrt{3x-2}-4\right)}{1+\sqrt{3x-2}}+\dfrac{\sqrt{x+3}-3}{2+\sqrt{x+3}}+4x-24=0\\
\leftrightarrow \dfrac{9\left(x-6\right)}{\left(1+\sqrt{3x-2}\right)\left(\sqrt{3x-2}+4\right)}+\dfrac{x-6}{\left(2+\sqrt{x+3}\right)\left(\sqrt{x+3}+3\right)}+4\left(x-6\right)=0\\
\to \left[\begin{array}{l}x=6\to y=10\\\dfrac{9}{\left(1+\sqrt{3x-2}\right)\left(\sqrt{3x-2}+4\right)}+\dfrac{1}{\left(2+\sqrt{x+3}\right)\left(\sqrt{x+3}+3\right)}+4=0\left(\text{vô nghiệm}\right)\end{array}\right.
\end{array}\]
Vậy \(\left(x;y\right)=\left\{\left(6;10\right);\left(1;26\right);\left(1;5\right)\right\}\)
