\(\text{Đặt: }A=-x^2-y^2+xy+2x+2y.\)

    \(\Rightarrow2A=-2x^2-2y^2+2xy+4x+4y=-\left(x^2-4x+4\right)-\left(y^2-y+4\right)-\left(x^2-2xy+y^2\right)+8\)

\(=8-\left(x-2\right)^2-\left(y-2\right)^2-\left(x-y\right)^2\)

(x²/2 - xy + y² /2) + (x² /2 - 2x + 2) + (y² /2 - 2y + 2) - 4 = (x/√2 - y √2)² + (x/√2 - √2)² + (y/√2 - √2)² - 4 >= -4

Vậy GTNN là -4 đạt được khi x = y = 2

Tìm GTNN chớ bạnbạn

vậy bạn giải cách GTNN cho mình xem đi được không 

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

\(\Rightarrow-2B=2x^2+2y^2-2xy-2x-4y\)

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

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

vì \(\left(x-y\right)^2\ge0\forall x,y;\left(x-2\right)^2\ge0\forall x;\left(y-2\right)\ge0\forall y\)nên

\(-2B=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge8\)

hay \(-2B\ge-8\Rightarrow B\le4\)

\(\Rightarrow maxB=4\Leftrightarrow\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=2\\y=2\end{cases}}}\)

-M = x^2+y^2-xy-2x-2y

-4M = 4x^2+4y^2-4xy-8x-8y

      = [ (4x^2-4xy+y^2) - 2.(2x-y).2 + 4 ] + (3y^2-12y+12)-16

      = [ (2x-y)^2 - 2.(2x-y).2 + 4 ] + 3.(y^2-4y+4) - 16

      = (2x-y-2)^2 + 3.(y-2)^2 - 16 >= -16 => M < = 4

Dấu "=" xảy ra <=> 2x-y-2 = 0 và y-2 = 0 <=> x = y = 2

Vậy ............

Tk mk nha

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\frac{2x+y+z}{2}\)

cmtt => GTLN

Tìm max:

Ta có:

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+xz}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

\(\le\frac{2x+y+z}{2}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\sqrt{2y+zx}\le\frac{2y+z+x}{2}\left(2\right)\\\sqrt{2z+xy}\le\frac{2z+x+y}{2}\left(3\right)\end{cases}}\)

Cộng (1), (2), (3) vế theo vế ta được

\(A\le\frac{2x+y+z}{2}+\frac{2y+z+x}{2}+\frac{2z+x+y}{2}=2\left(x+y+z\right)=4\)

Dấu = xảy ra khi \(x=y=z=\frac{2}{3}\)

Tìm min:

Ta có: \(\hept{\begin{cases}\sqrt{2x+yz}\ge0\\\sqrt{2y+zx}\ge0\\\sqrt{2z+xy}\ge0\end{cases}}\)

\(\Rightarrow A\ge0\)

Dấu = xảy ra khi \(\left(x,y,z\right)=\left(-2,2,2;2,-2,2;2,2,-2\right)\)

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{1}{2}\left(x+y+x+z\right)=\dfrac{1}{2}\left(2x+y+z\right)\)

Tương tự: \(\sqrt{2y+xz}\le\dfrac{1}{2}\left(x+2y+z\right)\) ; \(\sqrt{2z+xy}\le\dfrac{1}{2}\left(x+y+2z\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=4\)

\(P_{max}=4\) khi \(x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)

\(=\sqrt{3.\left(4+xy+yz+zx\right)}\)

Đã biết x² + y² + z² \(\ge\)xy + yz + zx

=> xy + yz + zx \(\le\dfrac{\left(x+y+z\right)^2}{3}\)

Khi đó \(P\le\sqrt{3\left(4+xy+yz+zx\right)}\le\sqrt{3\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}\)

= 4 

Dấu "=" xảy ra <=> x = 2/3