Giải thích các bước giải:

Ta có:

$\dfrac{2z-4x}3=\dfrac{3x-2y}4=\dfrac{4y-3z}2$

$\to \dfrac{6z-12x}9=\dfrac{12x-8y}{16}=\dfrac{8y-6z}4=\dfrac{6z-12x+12x-8y+8y-6z}{9+16+4}=0$

$\to 12x=8y=6z$

$\to 6x=4y=3z$

$\to 6x:12=4y:12=3z:12$

$\to \dfrac{x}2=\dfrac{y}3=\dfrac{z}4=k$

$\to x=2k, y=3k, z=4k, k\in N^*$ vì $x,y,z\in N^*$

Do $200<y^2+z^2<300$

$\to 200<(3k)^2+(4k)^2<300$

$\to 200<25k^2<300$

$\to 8<k^2<12$

$\to 2\sqrt2<k<2\sqrt3$

$\to k=3$ vì $k\in Z$

$\to x+y+z=2k+3k+4k=9k=9\cdot 3=27$