Question

Cho \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)   Chứng minh rằng \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

Answer 1

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

\(\Rightarrow\frac{3xz-2yz}{4z}=\frac{2yz-4xy}{3y}=\frac{4xy-3xz}{2x}\)

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

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

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

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

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

Answer 2

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Answer 3

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

\(\Rightarrow\frac{3xz-2yz}{4z}=\frac{2zy-4xy}{3y}=\frac{4yx-3zx}{2x}\)

\(\Rightarrow\frac{3xz-2yz+2zy-4xy+4yx-3zx}{4z+3y+2x}\)\(=\frac{0}{4z+3y+2x}=0\)

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

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

\(Từ\left(1\right),\left(2\right)\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\left(đpcm\right)\)