Câu hỏi của Dương Thanh Ngân

Question

Cho $x,y,z\in R$ thỏa mãn:

$\frac{3x^2}{2}+y^2+z^2+yz\:=1$

Tìm Min và Max của $B=x+y+z$

Nguyễn Việt Lâm · Accepted Answer

$3x^2+2y^2+2z^2+2yz=2$

$\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2zx+z^2\right)=2$

$\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2$

$\Leftrightarrow\left(x+y+z\right)^2=2-\left(x-y\right)^2-\left(x-z\right)^2\le2$

$\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}$

$B_{min}=-\sqrt{2}$ khi $\left\{{}\begin{matrix}x-y=0\x-z=0\x+y+z=-\sqrt{2}\end{matrix}\right.$ $\Rightarrow x=y=z=-\frac{\sqrt{2}}{3}$

$B_{max}=\sqrt{2}$ khi $x=y=z=\frac{\sqrt{2}}{3}$Câu trả lời: $3x^2+2y^2+2z^2+2yz=2$

$\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2zx+z^2\right)=2$

$\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2$

$\Leftrightarrow\left(x+y+z\right)^2=2-\left(x-y\right)^2-\left(x-z\right)^2\le2$

$\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}$

$B_{min}=-\sqrt{2}$ khi $\left\{{}\begin{matrix}x-y=0\x-z=0\x+y+z=-\sqrt{2}\end{matrix}\right.$ $\Rightarrow x=y=z=-\frac{\sqrt{2}}{3}$

$B_{max}=\sqrt{2}$ khi $x=y=z=\frac{\sqrt{2}}{3}$

Violympic toán 8