Question

[Jin Yi Hae]

giúp mk na m.n

Answer 1

[Nguyễn Huy Tú]

Giải:

Ta có: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\Rightarrow\left[\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)

Mà \(2x^2+2y^2-3z^2=-100\)

\(\Rightarrow18k^2+32k^2-75k^2=-100\)

\(\Rightarrow-25k^2=-100\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=2\)

+) \(k=2\Rightarrow x=6;y=8;z=10\)

\(\Rightarrow\left(x+y+z\right)^2=576\)

+) \(k=-2\Rightarrow x=-6;y=-8;z=-10\)

\(\Rightarrow\left(x+y+z\right)^2=576\)

Vậy \(\left(x+y+z\right)^2=576\)