❤ѕѕѕσиɢσкυѕѕѕ❤

Ta có: \(x^2+y^2+z^2+t^2-\left(xy+yz+zt+tx\right)=1-1\)

\(\Leftrightarrow2\left(x^2+y^2+z^2+t^2-xy-yz-zt-tx\right)=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2+2t^2-2xy-2yz-2zt-tx=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zt+t^2\right)+\left(t^2-2tx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-t\right)^2+\left(t-x\right)^2=0\)

Vì \(\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0;\left(z-t\right)^2\ge0;\left(t-x\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-t\right)^2+\left(t-x\right)^2\ge0\)

Dấu "=" xảy ra khi x - y = 0 ; y - z = 0 ; z - t = 0 ; t - x = 0 <=> x = y = z = t

Khi đó \(x^2+y^2+z^2+t^2=x^2+x^2+x^2+x^2=4x^2=1\)

\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)

Vậy \(x=y=z=t=\pm\frac{1}{2}\)

giúp mk vs

- x.y=-2; xz=3 =>x²yz=-2.3=-6

=>x²=\(\frac{-6}{yz}\) = -6/-4=2/3
- xz=3;yz=-4 => z²xy=3.-4=-12

=> z²=-12/xy=-12/-2=6
- xy=-2;yz=-4=>y²xz=-2.-4=8

=>y^2=8/xz=8/-4=-2

====>x²+y²+z²⁼2/3+6-2=14/3
 

ta có \(xy\le\left(\frac{x+y}{2}\right)^2\) và \(yz+xz=z\left(x+y\right)\le\frac{z^2+\left(x+y\right)^2}{2}\)

\(\Rightarrow5=xy+yz+xz\le\left(\frac{x+y}{2}\right)^2+\frac{z^2+\left(x+y\right)^2}{2}=\frac{3}{4}\left(x+y\right)^2+\frac{1}{2}z^2\)

Xét \(3x^2+3y^2+z^2\ge\frac{3}{2}\left(x+y\right)^2+z^2=2\left(\frac{3}{4}\left(x+y\right)^2+\frac{1}{2}z^2\right)\ge2\cdot5=10\)

dấu "=" xảy ra khi \(\hept{\begin{cases}x=y\\z=x+y\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=\pm1\\z=\pm2\end{cases}}}\)