\(\left\{{}\begin{matrix}x^3-3x^2+2=y\sqrt{y+3}\left(1\right)\\3\sqrt{x-2}+\sqrt{y-1}+x^2=12\left(2\right)\end{matrix}\right.\) \(\left(x\ge2;y\ge1\right)\)
\(\left(1\right)\Leftrightarrow\left(x^3-3x^2+3x-1\right)-\left(3x-3\right)=\left(y+3-3\right)\sqrt{y+3}\)
\(\Leftrightarrow\left(x-1\right)^3-3\left(x-1\right)=\sqrt{y+3}^3-3\sqrt{y+3}\)
Đặt \(x-1=a;\sqrt{y+3}=b\left(a\ge1;b\ge2\right)\)
\(\Rightarrow a^3-3a=b^3-3b\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2-3\right)=0\)
\(\Rightarrow a=b\left(\text{vì }a^2+ab+b^2-3\ge4>0\right)\)
\(\Rightarrow x-1=\sqrt{y+3}\)
\(\Rightarrow x^2-2x+1=y+3\)
\(\Leftrightarrow y-1=x^2-2x-3\). Thay vào (2)
\(\Rightarrow3\sqrt{x-2}+\sqrt{x^2-2x-3}+x^2=12\)
\(\Leftrightarrow3\sqrt{x-2}-3+\sqrt{\left(x-3\right)\left(x+1\right)}+x^2-9=0\)
\(\Leftrightarrow\dfrac{3\left(x-2-1\right)}{\sqrt{x-2}+1}+\sqrt{\left(x-3\right)\left(x+1\right)}+\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\sqrt{x-3}\left[\dfrac{3\sqrt{x-3}}{\sqrt{x-2}+1}+\sqrt{x+1}+\sqrt{x-3}\left(x+3\right)\right]=0\)
\(\Rightarrow x=3\)
\(\Rightarrow y=1\)
Vậy hệ có nghiệm duy nhất (x ; y) = (3 ; 1)
