Ap dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=0\)
\(\Leftrightarrow12x-8y=0;8y-6z=0;6z-12x=0\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4};\frac{z}{4}=\frac{x}{2};\frac{y}{3}=\frac{z}{4}\)
Vậy : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) (dqcm)
(3x-2y)/4 = (2z-4x)/3 = (4y-3z)/2 =
= (12x-8y)/16 = (6z-12x)/9 = (8y-6z)/4 = (12x-8y + 6z-12x + 8y-6z)/(16+9+4) = 0
<=>
{12x - 8y = 0
{6z - 12x = 0
{8y - 6z = 0
<=>
{x/2 = y/3
{z/4 = x/2
{y/3 = z/4
<=> x/2 = y/3 = z/4( đpcm)
