Question

Tìm x,y.z : \(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\) và 3x - 2y + 5z = 96

Answer 1

\(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\)

\(\Rightarrow\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}=\dfrac{5z-6y+6x-4z+4y-5x}{4+5+6}=\dfrac{x-2y+z}{4+5+6}\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{-2y}{5}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{3x}{12}=\dfrac{-2y}{5}=\dfrac{5z}{30}\)

\(\Rightarrow\dfrac{3x}{12}=\dfrac{-2y}{5}=\dfrac{5z}{30}=\dfrac{3x-2y+5z}{12-5+30}=\dfrac{96}{37}\)