Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{16+9+4}=0\)

\(\Rightarrow3x-2y=0\)

\(\Rightarrow2z-4x=0\)

\(\Rightarrow4y-3z=0\)

Ta có: \(3x-2y=0\Rightarrow3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\) (1)

\(2z-4x=0\Rightarrow2z=4x\Rightarrow\frac{z}{4}=\frac{x}{2}\) (2)

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

\(\Rightarrowđpcm\)

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

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

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

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\frac{0}{29}=0\)

\(\Rightarrow\begin{cases}12x-8y=0\\6z-12x=0\\8y-6z=0\end{cases}\)\(\Rightarrow\begin{cases}12x=8y\\6z=12x\\8y=6z\end{cases}\)\(\Rightarrow12x=8y=6z\)

\(\Rightarrow\frac{12x}{24}=\frac{8y}{24}=\frac{6z}{24}\)

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

CHTT đầy

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

\(\Rightarrow\frac{4\left(3x-2y\right)}{4.4}=\frac{3\left(2z-4x\right)}{3.3}=\frac{2\left(4y-3z\right)}{2.2}\)

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

Áp dụng TCDTSBN ta có:

\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)

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

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

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

ta có:\(\frac{3x-2y}{4}\)=\(\frac{2z-4x}{3}\)=\(\frac{4y-3z}{2}\)
       =\(\frac{12x-8y}{16}\)=\(\frac{6z-12x}{9}\)=\(\frac{8y-6z}{4}\)=\(\frac{12x-8y+6z-12x+8y-6z}{16+9+4}\)
=>\(\hept{\begin{cases}3x-2y=0\\2z-4x=0\\4y-3z=0\end{cases}}\)=>\(\hept{\begin{cases}3x=2y\\2z=4x\\4y-3z\end{cases}}\)=>\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{z}{4}=\frac{x}{2}\\\frac{y}{3}=\frac{z}{4}\end{cases}}\)=>\(\hept{\begin{cases}x\\2\end{cases}}=\frac{y}{3}=\frac{z}{4}\)

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

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

\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\frac{0}{29}=0\)

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

\(\Rightarrow\frac{4y-3z}{2}=0\Rightarrow4y-3z=0\Rightarrow4y=3z\Rightarrow\frac{y}{3}=\frac{z}{4}\)( 2 )

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

Ta có :

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

suy ra : \(\frac{4.\left(3x-2y\right)}{16}=\frac{3.\left(2z-4x\right)}{9}=\frac{2.\left(4y-3z\right)}{4}\)

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

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

\(\frac{2z-4x}{3}=0\Rightarrow2z=4x\Rightarrow\frac{x}{2}=\frac{z}{4}\)( 2 )

Từ ( 1 ) và ( 2 ) ta được : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\)\(\frac{6z-12x}{9}=\frac{8y-6z}{4}\)\(=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)

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

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

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

~~ học tốt ~~

\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{3xz-2yz}{4z}=\frac{2yz-4xy}{3y}=\frac{4xy-3xz}{2x}=\frac{\left(3xz-2yz\right)+\left(2yz-4xy\right)+\left(4xy-3xz\right)}{4z+3y+2x}=0\)

\(\Rightarrow\hept{\begin{cases}3x-2y=0\\2z-4x=0\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\2z=4x\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{x}{2}=\frac{z}{4}\end{cases}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}}\)

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