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

 = (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 

cần tính x,y,z

làm đến đây rồi apf dụng tính chất đúng ko

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:

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

b tham khảo nhé

làm xong mà vào viện ah

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Ok