Question

Cho các số x, y, z thỏa mãn \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3x}{2}\)

Tính giá trị của biểu thức \(A=\dfrac{2x+2y+20z}{x+y+z}\)

Answer 1

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{2x}\\ \Rightarrow\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{2x}=\dfrac{\left(3xz-3xz\right)+\left(2yz-2yz\right)+\left(4xy-4xy\right)}{4z+3y+2x}=0\\ \Rightarrow3x-2y=2z-4x=4y-3z=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\)

3x=2y => \(\dfrac{x}{2}=\dfrac{y}{3}\)

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

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

Thế dô A ; tự tinh !!