Câu hỏi của Nguyễn Đăng Nhân

Question

Tìm GTLN của:

$A=-x^2+2xy-4y^2+2x+10y-3$

Lê Song Phương · Accepted Answer

Ta có $A=-x^2+2xy-4y^2+2x+10y-3$

$A=-x^2+2\left(y+1\right)x-4y^2+10y-3$

$A=-x^2+2\left(y+1\right)x-\left(y+1\right)^2-3y^2+12y-2$

$A=-\left[x-\left(y+1\right)\right]^2-3\left(y^2-4y+4\right)+10$

$A=-\left(x-\left(y+1\right)\right)^2-3\left(y-2\right)^2+10$ $\le10$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}x=y+1\y-2=0\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(3,2\right)$

Vậy $max_A=10$Câu trả lời: Ta có $A=-x^2+2xy-4y^2+2x+10y-3$

$A=-x^2+2\left(y+1\right)x-4y^2+10y-3$

$A=-x^2+2\left(y+1\right)x-\left(y+1\right)^2-3y^2+12y-2$

$A=-\left[x-\left(y+1\right)\right]^2-3\left(y^2-4y+4\right)+10$

$A=-\left(x-\left(y+1\right)\right)^2-3\left(y-2\right)^2+10$ $\le10$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}x=y+1\y-2=0\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(3,2\right)$

Vậy $max_A=10$

đào bảo · Answer

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