Question

tim GTLN cua cacda thuc sau:

1)A=-4x^2-5y^2+8xy+10y+12               2)B=-x^2-y^2+xy+2x+2y

3)C=-x^2+2xy-4y^2+2x+10y-3             4)D=-x^2+xy-y^2-2x+4y+11

5)E=x+y+z-(x^2+2y^2+4z^2)                6)F=-x^2-y^2-z^2+xy+yz+zx-3

Answer 1

\(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4\left(x^2-2xy+y^2\right)-\left(y^2-10y+25\right)+37\)

\(=-4\left(x-y\right)^2-\left(y-5\right)^2+37\le37\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-5\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=5\)

Vậy \(MaxA=37\)

Answer 2

\(B=-x^2-y^2+xy+2x+2y\)

\(=\dfrac{-2x^2-2y^2+2xy+4x+4y}{2}\)

\(=\dfrac{-\left(x^2-4x+4\right)-\left(y^2-4y+4\right)-\left(x^2-2xy+y^2\right)+8}{2}\)

\(=\dfrac{-\left(x-2\right)^2-\left(y-2\right)^2-\left(x-y\right)^2+8}{2}\le\dfrac{8}{2}=4\)

Dấu "=" xảy ra \(\Leftrightarrow...\Leftrightarrow x=y=2\)

Vậy \(MaxB=4\)

Answer 3

\(C=-x^2+2xy-4y^2+2x+10y-3\)

\(=-\left(x^2+y^2+1-2x+2y-2xy\right)-3\left(y^2-4y+4\right)-3-1+12\)

\(=-\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]-3\left(y-2\right)^2+8\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+8\le8\)

Dấu "=" xảy ra \(\Leftrightarrow...\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Vậy \(MaxC=8\)

Answer 4

\(F=-x^2-y^2-z^2+xy+yz+zx-3\)

\(=\dfrac{-2x^2-2y^2-2z^2+2xy+2yz+2zx-6}{2}\)

\(=\dfrac{-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2-6}{2}\le-\dfrac{6}{2}=-3\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

Vậy \(MaxF=-3\)

Answer 5

Dự đoán điểm rơi: \(x=0,y=2\)

\(D=-x^2+xy-y^2-2x+4y+11\)

\(=\dfrac{-2x^2+2xy-2y^2-4x+8y+22}{2}\)

\(=\dfrac{-\left(x^2+y^2+4-2xy+4x-4y\right)-x^2-\left(y^2-4y+4\right)+22+4+4}{2}\)

\(=\dfrac{-\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]-x^2-\left(y-2\right)^2+30}{2}\)

\(=\dfrac{-\left(x-y+2\right)^2-x^2-\left(y-2\right)^2+30}{2}\le\dfrac{30}{2}=15\)

Dấu "=" xảy ra \(\Leftrightarrow...\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

Vậy \(MaxD=15\)

Answer 6

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

\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)-2\left(y^2-\dfrac{1}{2}y+\dfrac{1}{16}-\dfrac{1}{16}\right)-4\left(z^2-\dfrac{1}{4}z+\dfrac{1}{64}-\dfrac{1}{64}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}-2\left(y-\dfrac{1}{4}\right)^2+\dfrac{1}{8}-4\left(z-\dfrac{1}{8}\right)^2+\dfrac{1}{16}\le\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}=\dfrac{7}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow...\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{4}\\z=\dfrac{1}{8}\end{matrix}\right.\)

Vậy \(MaxE=\dfrac{7}{16}\)