Giải hệ phương trình \(\left\{\begin{matrix} x^3-y^3+3x-12y+7=3x^2-6y^2

Điều kiện: \(\left\{\begin{matrix} x+2\geq 0\\ 4-y\geq 0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ y\leq 4 \end{matrix}\right.\)

Từ phương trình (1) ta có \((x-1)^3=(y-2)^2\Leftrightarrow x-1=y-2\Leftrightarrow y=x+1\)    (3)
Thay (3) vào (2) ta được pt: \(\sqrt{x+2}+\sqrt{4-(x+1)}=x^3+(x+1)^2-4x-2(x+1)\)

\(\Leftrightarrow \sqrt{x+2}+\sqrt{3-x}=x^3+x^2-4x-1, \ dk -2\leq x\leq 3\)

\(\Leftrightarrow (\sqrt{x+2}+\sqrt{3-x})-3=x^3+x^2-4x-4\)
\(\Leftrightarrow \frac{2(\sqrt{(x+2)(3-x)}-2)}{(\sqrt{x+2}+\sqrt{3-x})+3}=(x+1)(x^2-4)\)
\(\Leftrightarrow \frac{2\left [ (x+2)(3-x)-4 \right ]}{(\sqrt{x+2}+\sqrt{3-x}+3)(\sqrt{(x+2)(3-x)}+2)}=(x+1)(x^2-4)\)
\(\Leftrightarrow \frac{2(-x^2+x+2)}{(\sqrt{x+2}+\sqrt{3-x}+3)(\sqrt{(x+2)(3-x)}+2)}=(x+2)(x^2-x-2)\)
\(\small \Leftrightarrow (x^2-x-2)\left ( \underbrace{x+2+ \frac{2}{(\sqrt{x+2}+\sqrt{3-x}+3)(\sqrt{(x+2)(3-x)}+2)} }_{>0} \right )=0\)

\(x^2-x-2=0\Leftrightarrow x=2 \ \vee \ x=-1\)

• \(x=2\overset{(3) }{\rightarrow}y=3\Rightarrow (x;y)=(2;3)\) (thỏa mãn đ/k)
• \(x=-1\overset{(3) }{\rightarrow}y=0\Rightarrow (x;y)=(-1;0)\)  (thỏa mãn đ/k)

Vậy hệ phương trình có hai nghiệm ( x; y) = (2;3),( x; y ) = (-1;0)