Bài 1:
Ta có: \(A=2022-x^2-10y^2-6xy+4y\)
\(=-\left(-2022+x^2+10y^2+6xy-4y\right)\)
\(=-\left(x^2+6xy+9y^2+y^2-4y+4-2026\right)\)
\(=-\left[\left(x^2+6xy+9y^2\right)+\left(y^2-4y+4\right)-2026\right]\)
\(=-\left(x+3y\right)^2-\left(y-2\right)^2+2026\)
\(=-\left[\left(x+3y\right)^2+\left(y-2\right)^2\right]+2026\)
Ta có: \(\left(x+3y\right)^2\ge0\forall x,y\)
\(\left(y-2\right)^2\ge0\forall y\)
Do đó: \(\left(x+3y\right)^2+\left(y-2\right)^2\ge0\forall x,y\)
\(\Leftrightarrow-\left[\left(x+3y\right)^2+\left(y-2\right)^2\right]\le0\forall x,y\)
\(\Leftrightarrow-\left[\left(x+3y\right)^2+\left(y-2\right)^2\right]+2026\le2026\forall x,y\)
Dấu '=' xảy ra khi:
\(\left\{{}\begin{matrix}x+3y=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3\cdot2=0\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=2\end{matrix}\right.\)
Vậy: Giá trị lớn nhất của biểu thức \(A=2022-x^2-10y^2-6xy+4y\) là 2026 khi x=-6 và y=2
