Theo bài ra ta có:
\(\dfrac{2x-3y}{2}=\dfrac{4y-2z}{3}=\dfrac{3z-4x}{4}=\dfrac{8x-12y}{8}=\dfrac{12y-6z}{9}=\dfrac{6z-8x}{8}\)
\(=\dfrac{8x-12y+12y-6z+6z-8x}{8+9+8}=\dfrac{0}{25}=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-3y=0\Leftrightarrow2x=3y\Leftrightarrow x=\dfrac{3}{2}y\\4y-2z=0\Leftrightarrow4y=2z\Leftrightarrow z=2y\\\end{matrix}\right.\)
Vậy 3x+2y+z=\(\dfrac{3}{2}y.3+2y+2y=y\left(\dfrac{9}{2}+2+2\right)=\dfrac{17}{2}y=17\Rightarrow y=2\Rightarrow x=3;z=4\)
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