Question

Cho \(\frac{3x-2y}{4}\) = \(\frac{2z-4x}{3}\) = \(\frac{4y-3z}{2}\) .CMR : \(\frac{x}{2}\) = \(\frac{y}{3}\) =\(\frac{z}{4}\)

Answer 1

ta có; \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)

theo t/c dãy tỉ số bằng nhau ta có: \(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=0\)

do đó: 12x-8y=0 =>\(\frac{x}{2}=\frac{y}{3}\)

6z-12x=0 => \(\frac{z}{4}=\frac{x}{2}\)

8y-6z=0 => \(\frac{y}{3}=\frac{z}{4}\)

vậy \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)