Ta có \(3x^2+4y^2+6x+3y-4=0\Rightarrow3x^2+6x+3+4\left(y^2+\dfrac{3}{4}y-\dfrac{7}{4}\right)=0\Rightarrow3\left(x+1\right)^2+4\left[\left(y+\dfrac{3}{8}\right)^2-\dfrac{121}{64}\right]=0\Rightarrow3\left(x+1\right)^2+\left(2y+\dfrac{3}{4}\right)^2=\dfrac{121}{16}\Rightarrow3\left(x+1\right)^2=\dfrac{121}{16}-\left(2y+\dfrac{3}{4}\right)^2\le\dfrac{121}{16}\Rightarrow\left(x+1\right)^2\le\dfrac{121}{48}< 3\)Do đó \(\left(x+1\right)^2\in\left\{0,1\right\}\)
-Xét \(\left(x+1\right)^2=0\Rightarrow x=-1\).Thay vào, ta có y = 1.
-Xét \(\left(x+1\right)^2=1\Rightarrow x+1\in\left\{1,-1\right\}\Rightarrow x\in\left\{0,-2\right\}\).Thay vào, ta có y ∈ ∅
