\(5x^2+2y^2+6xy-8x-4y+4=0\)
\(\Leftrightarrow4x^2+x^2+y^2+y^2+2xy+4xy-8x-4y+4=0\)
\(\Leftrightarrow\left(4x^2+y^2+4+4xy-8x-4y\right)+\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left[\left(2x\right)^2+4xy+y^2-4\left(2x+y\right)+2^2\right]+\left(x+y\right)^2=0\)
\(\Leftrightarrow\left[\left(2x+y\right)^2-2\cdot\left(2x+y\right)\cdot2+2^2\right]+\left(x+y\right)^2=0\)
\(\Leftrightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(2x+y-2\right)^2\ge0\forall x,y\\\left(x+y\right)^2\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2\ge0\forall x,y\)
Mặt khác: \(\left(2x+y-2\right)^2+\left(x+y\right)^2=0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2x+y-2=0\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot\left(-y\right)+y-2=0\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y+y-2=0\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=2\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
Thay x,y vào P ta có:
\(P=2^{2023}+\left(-2\right)^{2023}=2^{2023}-2^{2023}=0\)
Vậy: ...
