\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
=>\(\dfrac{4\left(3x-2y\right)}{4.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)
=>\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
=>\(\dfrac{12x-8y}{16}=0\)
=>12x-8y=0
=>12x=8y
=>\(\dfrac{12x}{24}=\dfrac{8y}{24}\)
=>\(\dfrac{x}{2}=\dfrac{y}{3}\)(1)
Lại có \(\dfrac{8y-6z}{4}=0\)
=>8y-6z=0
=>8y=6z
=>\(\dfrac{8y}{24}=\dfrac{6z}{24}\)
=>\(\dfrac{y}{3}=\dfrac{z}{4}\)(2)
từ (1) và (2)=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
