Question

[Nguyễn Tuấn Hưng]

 Tìm giá trị lớn nhất của biểu thức:

A=-x²-5y²+2xy-4x+20y+13

B=-7x²-y²+4xy+16x-2y+17

Answer 1

[Phạm Thị Thùy Linh]

\(A=-x^2-5y^2+2xy-4x+20y+13\)

\(=-x^2+2xy-y^2-4y^2-4x+4y+16y+13\)

\(=-\left(x^2-2xy+y^2\right)-\left(4y^2-16y+16\right)-\left(4x-4y\right)+29\)

\(=-\left(x-y\right)^2-4\left(y-2\right)^2-4\left(x-y\right)-4+25\)

\(=-\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]-4\left(y-2\right)^2+25\)

\(=-\left(x-y+2\right)^2-4\left(y-2\right)^2+25\)

\(A_{max}=25\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y+2=0\\y=2\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)

Answer 2

[Phạm Thị Thùy Linh]

\(B=-7x^2-y^2+4xy+16x-2y+17.\)

\(=-4x^2+4xy-y^2-3x^2+12x-12+4x-2y+29\)

\(=-\left(2x-y\right)^2-3\left(x-2\right)^2+2\left(2x-y\right)^2-1+30\)

\(=-\left[\left(2x-y\right)^2-2\left(2x-y\right)^2+1\right]-3\left(x-2\right)^2+30\)

\(=-\left(2x-y-1\right)^2-3\left(x-2\right)^2+30\)

\(\Rightarrow B_{max}=30\Leftrightarrow\hept{\begin{cases}\left(2x-y-1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-y-1=0\\x=2\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)