Question

cho x,y,z thỏa mãn 0<=x,y,z<=2 

x+y+z=3

xác định giá trị lớn nhất của A=x^2+y^2+z^2

Answer 1

\(\Rightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\le0\)

\(\Leftrightarrow xyz-2\left(xy+yz+xz\right)+4\left(x+y+z\right)-8\le0\)

\(\Leftrightarrow-2\left(xy+yz+xz\right)\le8-4\left(x+y+z\right)-xyz=8-4.3+0=-4\left(xyz\ge0\right)\)

\(A=x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+xz\right)\le3^2-4=5\)

\(max_A=5\Leftrightarrow\left\{{}\begin{matrix}xyz=0\\\left(x-2\right)\left(y-2\right)\left(z-2\right)=0\\x+y+z=3\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y;z\right)=\left\{0;1;2\right\}\) \(và\) \(các\) \(hoán\) \(vị\)