a) \(\dfrac{3cy-4bz}{2x}=\dfrac{4az-2cx}{3y}=\dfrac{2bx-3ay}{4z}\)
=> \(\dfrac{3cy-4bz}{2x}.\dfrac{2x}{2x}=\dfrac{4az-2cx}{3y}.\dfrac{3y}{3y}=\dfrac{2bx-3ay}{4z}.\dfrac{4z}{4z}\)
=> \(\dfrac{6cxy-8bzx}{4x^2}=\dfrac{12azy-6cxy}{9y^2}=\dfrac{8bxz-12ayz}{16z^2}\)
Áp dụng t/c ...
\(\dfrac{6cxy-8bzx}{4x^2}=\dfrac{12azy-6cxy}{9y^2}=\dfrac{8bxz-12ayz}{16z^2}=\dfrac{6cxy-8bzx+12azy-6cxy+8bxz-12ayz}{4x^2+9y^2+16z^2}=\dfrac{0}{4x^2+9y^2+16z^2}=0\)
Ta có : 6cxy - 8bzx = 0
=> 6cxy = 8bzx
=>3cx = 4bz
=>\(\dfrac{c}{4z}=\dfrac{b}{3y}\) (1)
Ta có : 12azy - 6cxy = 0
=> 12azy = 6cxy
=> 4az = 2cx
=> \(\dfrac{a}{2x}=\dfrac{c}{4z}\) (2)
Từ (1),(2) => \(\dfrac{a}{2x}=\dfrac{b}{3y}=\dfrac{c}{4z}\) (ĐPCM)
