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mơn bb*

\(A=-x^2-4xy-5y^2-6y+1672\)

\(=-x^2-4xy-4y^2-y^2-6y-9+1681\)

\(=-\left(x^2+4xy+4y^2\right)-\left(y^2+6y+9\right)+1681\)

\(=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\)

Dễ thấy: \(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\forall x,y\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}-\left(x+2y\right)^2\le0\forall x,y\\-\left(y+3\right)^2\le0\forall y\end{matrix}\right.\)

\(\Rightarrow-\left(x+2y\right)^2-\left(y+3\right)^2\le0\forall x,y\)

\(\Rightarrow A=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\le1681\forall x,y\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}-\left(x+2y\right)^2=0\\-\left(y+3\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+2y=0\\y+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2y\\y=-3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=-3\end{matrix}\right.\)

Ta có :

\(D=-x^2-4xy-5y^2+6y+1672\)

\(=-\left(x^2+4xy+4y^2\right)-\left(y^2+6y+9\right)+9+1672\)

\(=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\)

Có :

\(\left(x+2y\right)^2\ge0\)

\(\left(y+3\right)^2\ge0\)

\(\Rightarrow-\left(x+2y\right)^2-\left(y+3\right)^2+1681\le1681\)

\(\Rightarrow Max_N=1681\Leftrightarrow\hept{\begin{cases}y=-3\\x=6\end{cases}}\)

Vậy ...

-(x-2y)^2 -(y-3)^2 +1681

Với mọi x, y ta có: -....<=0

=>-.... <= 1681

Dấu = xảy ra khi

x=2y; y=3

=> x=6;y=3

Vậy...

Trần Thủy Dung làm sai rồi +6y=-(-6y) chứ

a) A = x² - 6x + 13 = x² - 2.x.3 + 3³ +4 = (x-3)^{2 +} 4 >= 4 suy ra minA=4 
mấy câu kia giải tương tự

\(C=1-6y-5y^2-12xy-9x^2\)

\(\Rightarrow C=-4y^2-12xy-9x^2-y^2-6y+1\)

\(\Rightarrow C=-\left(4y^2+12xy+9x^2\right)-\left(y^2+6y+9\right)+1+9\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\)

mà \(\left\{{}\begin{matrix}-\left(2y-3x\right)^2\le0,\forall x;y\\-\left(y+3\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\le10\)

\(\Rightarrow GTLN\left(C\right)=10\left(tạix=-2;y=-3\right)\)

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)