\(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}=\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2z-4x=0\\4y-3z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\\ \Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x-2y+3z}{2-6+12}=\dfrac{8}{8}=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\\z=4\end{matrix}\right.\)

Ta có \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2x-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}=\dfrac{12x-8y-12x+8y-6z}{29}\)

Do đó:

\(\dfrac{3x-2y}{4}=0\Rightarrow3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\left(1\right)\)

\(\dfrac{2z-4x}{3}=0\Rightarrow2z=4x\Rightarrow\dfrac{x}{2}=\dfrac{z}{4}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\). Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\Rightarrow x=4;y=6;z=8\)

Ta có \(\frac{3 x - 2 y}{4} = \frac{2 z - 4 x}{3} = \frac{4 y - 3 z}{2}\)

\(\Rightarrow \frac{4 \left(\right. 3 x - 2 y \left.\right)}{16} = \frac{3 \left(\right. 2 x - 4 x \left.\right)}{9} = \frac{2 \left(\right. 4 y - 3 z \left.\right)}{4} = \frac{12 x - 8 y - 12 x + 8 y - 6 z}{29}\)

\(\Rightarrow\frac{3 x - 2 y}{4}=0\Rightarrow3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\left(\right.1\left.\right)\)

\(\frac{2 z - 4 x}{3} = 0 \Rightarrow 2 z = 4 x \Rightarrow \frac{x}{2} = \frac{z}{4} \left(\right. 2 \left.\right)\)

Từ (1) và (2) suy ra \(\frac{x}{2} = \frac{y}{3} = \frac{z}{4}\).

Áp dụng ( hoặc theo) tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x + y + z}{2 + 3 + 4}=\frac{18}{9}=2\Rightarrow x=2\cdot2=4;y=2\cdot3=6;z=2\cdot4=8\)

b tham khảo nhé

Ai làm được thì giúp mình với ;-;

suy ra:

\(\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)

\(=\dfrac{12x-8y+6z-12x+8y-6z}{29}=0\)

Vậy

\(\dfrac{3x-2y}{4}=0\Rightarrow3x=\dfrac{2y\Rightarrow x}{2}=\dfrac{y}{3}\left(1\right)\)

\(\dfrac{2z-4x}{4}=0\Rightarrow2z=4x\Rightarrow\dfrac{x}{2}=\dfrac{z}{4}\left(2\right)\)

từ (1) và (2) ta được\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

suy ra:

$\frac{4\left(3x-2y\right)}{16}=\frac{3\left(2z-4x\right)}{9}=\frac{2\left(4y-3z\right)}{4}$

$=\frac{12x-8y+6z-12x+8y-6z}{29}$