XL gtnn B = 19/4

GTNN = -1/4

\(-2x^2-2xy-y^2+2x-2y-2=-\left[y^2+2y\left(x+1\right)+\left(x+1\right)^2\right]-\left(x^2-4x+4\right)+3=-\left(y+x+1\right)^2-\left(x-2\right)^2+3\le3\)

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

Học tốt!

`2(x^2+y^2)+z^2=-2xy+2yz-4x-4`

`<=>2x^2+2y^2+z^2+2xy-2yz+4x+4=0`

`<=>(x^2+2xy+y^2)+(y^2-2yz+z^2)+(x^2+4x+4)=0`

`<=>(x+y)^2+(y-z)^2+(x+2)^2=0`

Vì `VT>=0`

Nên dấu "=" xảy ra khi `x+y=0,y-z=0,x+2=0`

`<=>x=-y,y=z,x=-2`

`<=>x=-2,y=z=-x=2`

Vậy `(x,y,z)=(-2,2,2)`

c.ơn nhé !

cậu học giỏi quá nha !

Ta có: \(P=2x-2xy-2x^2-y^2\)

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

\(P=-\left(x^2+2xy+y^2\right)-\left(x^2-2x+1\right)+1\)

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

\(P=-\left[\left(x+y\right)^2+\left(x-1\right)^2\right]+1\le1\forall x;y\)

Vậy GTLN của P là 1 khi x=-1; y=1. 

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

cám ơn bạn nhá :)

\(=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)

\(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right).\left(x-y+3\right)\)