Bài 2 :

a) \(P=x^2+y^2+xy+x+y\)

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

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

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

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

\(P=\frac{\left(x+y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2}{2}-1\le-1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+1=0\\y+1=0\end{cases}}\)

Mình nghĩ đề phải là tìm GTLN của \(P=x^2+y^2+xy+x-y\)hoặc đổi dấu x và y thì dấu "=" mới xảy ra đc

@ Phương ơi ! Cái dòng \(P=\)cuối ấy . Chỗ đấy là \(\ge-1\)em nhé!

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

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 \(\left(x,y,z\right)=\left(-2,2,2;2,-2,2;2,2,-2\right)\)

\(x+2y=1\Rightarrow x=1-2y\)

a/ \(A=x^2+y^2=\left(1-2y\right)^2+y^2=5y^2-4y+1=5\left(y-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}\)

\(A_{min}=\frac{1}{5}\) khi \(\left\{{}\begin{matrix}x=\frac{1}{5}\\y=\frac{2}{5}\end{matrix}\right.\)

b/ \(B=\left(1-2y\right)y=-2y^2+y=-2\left(y-\frac{1}{4}\right)^2+\frac{1}{8}\le\frac{1}{8}\)

\(B_{max}=\frac{1}{8}\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)