Question

a, cho \(\dfrac{3cy-4bz}{2x}=\dfrac{4az-2cx}{3y}=\dfrac{2bx-3ay}{4z}\)

CMR: \(\dfrac{a}{2x}=\dfrac{b}{3y}=\dfrac{c}{4z}\)

b, Cho \(\dfrac{7cy-5bz}{x}=\dfrac{2az-7cx}{y}=\dfrac{5bx-2ay}{z}\)

CMR : \(\dfrac{2a}{x}=\dfrac{5b}{y}=\dfrac{7c}{z}\)

Answer 1

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)

Answer 2

À mà , phần b) tương tự nhé