\(x^2-4x+y^2-6y+15=2\)
\(\left(x^2-4x+4\right)+\left(y^2-6y+9\right)=0\)
\(\left(x-2\right)^2+\left(y-3\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow VT\ge0\forall x,y\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
1) \(x^2-4x+y^2-6y+15=2\)
\(\Leftrightarrow x^2-4x+y^2-6y+13=0\)
\(\Leftrightarrow x^2-4x+4+y^2-6y+9=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow x-2=0;y-3=0\)
\(\Leftrightarrow x=2;y=3\)
2) \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+x+1\right)+2x\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2+x+1=0\) hay \(\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\) (pt vô nghiệm) hay \(x=-1\)
-Vậy \(S=\left\{-1\right\}\)
