Đáp án:
đặt `x/7=y/3=z/2=k` `(k` khác `0)`
`<=> x=7k; y=3k; z=2k`
thay ` x=7k; y=3k; z=2k` vào bt `x^2+y^2-z^2=216` ta có:
`(7k)^2+(3k)^2-(2k)^2=216`
`<=> 49k^2+9k^2-4k^2=216`
`<=> 54k^2=216`
`<=> k^2=4`
`<=> k=+-2`
khi `k=2` thì
`x/7=2<=>x=2.7=14`
`y/3=2<=>y=2.3=6`
`z/2=2<=>z=2.2=4`
khi `k=-2` thì
`x/7=-2<=>x=-2.7=-14`
`y/3=-2<=>y=-2.3=-6`
`z/2=-2<=>y=-2.2=-4`
vậy `(x;y;z)=(14;6;4);(-14;-6;-4)`
