BẠN NÀO LÀM ĐÚNG MÌNH K NHA
\(A=-x^2+4xy-5y^2+6y-17\)
\(=-\left(x^2-4xy+4y^2\right)-\left(y^2-6y+9\right)-8\)
\(=-\left(x-2y\right)^2-\left(y-3\right)^2-8\)
Vì \(\hept{\begin{cases}-\left(x-2y\right)^2\le0;\forall x,y\\-\left(y-3\right)^2\le0;\forall x,y\end{cases}}\)
\(\Rightarrow-\left(x-2y\right)^2-\left(y-3\right)^2\le0;\forall x,y\)
\(\Rightarrow-\left(x-2y\right)^2-\left(y-3\right)^2-8\le0-8;\forall x,y\)
Hay \(A\le-8;\forall x,y\)
Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-2y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)
Vậy MAX \(A=-8\)\(\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)
\(A=-x^2+4xy-5y^2+6y-17\)
\(=-x^2+4xy-4y^2-y^2+6y-9-8\)
\(=-\left(x^2-4xy+4y^2\right)-\left(y^2-6y+9\right)-8\)
\(=-\left(x-2y\right)^2-\left(y-3\right)^2-8\le-8\)
Vậy GTLN của A = -8
\(\Leftrightarrow\hept{\begin{cases}x-2y=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)
Best_Suarez (✰Ƙ❤C✰)
Thiếu
\(A=-x^2+4xy-5y^2+6y-17\)
\(A=-\left(x^2-4xy+4y^2\right)-\left(y^2-6y+9\right)-8\)
\(A=-\left(x-2y\right)^2-\left(y-3\right)^2-8\le-8\)
Vì \(\hept{\begin{cases}-\left(x-2y\right)^2\le0\\-\left(y-3\right)^2\le0\end{cases}\forall x}\)nên \(-\left(x-2y\right)^2-\left(y-3\right)^2\le0\)
\(\Rightarrow-\left(x-2y\right)^2-\left(y-3\right)^2-8\le8\)
Dấu '' = '' xảy ra
\(\Leftrightarrow\hept{\begin{cases}x-2y=0\\y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-2y=0\\y=3\end{cases}\Leftrightarrow\hept{\begin{cases}x-6=0\\y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=3\end{cases}}}\)
