Tham khảo nhá

(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

Học tốt!

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

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

\(\Rightarrow\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\Rightarrow\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)

\(=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}\)

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

\(=\frac{0}{16+9+4}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2z-4x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=2y\\2z=4x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{3}\\\frac{x}{2}=\frac{z}{4}\end{matrix}\right.\)

\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\left(đpcm\right)\)

Số nào là số lẻ : 2 78 467 1356 13464 368634 4580744 56767533

Ta có: (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 

Vậy<=> x/2 = y/3 = z/4 

Quá đúng Đặng Phương Thảo ơi

cậu quên chưa xét th x=y=z=0

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

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

\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

Dựa vào tính chất dãy tỉ số bằng nhau ta có:

\(=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}\)

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

\(\Rightarrow\left\{{}\begin{matrix}12x=8y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\\6z-12x=0\Rightarrow\dfrac{x}{2}=\dfrac{z}{4}\\8y=6z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

(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

Học tốt!

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