Question

Bài 6

Cho : \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)

Tính giá trị của :

\(E=\dfrac{x+2y+3\text{z}}{x-2y+3\text{z}}\) ( với x - 2y - 3z khác 0)

Answer 1

Theo tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{5}\) = \(\dfrac{y}{4}\) = \(\dfrac{z}{3}\) = \(\dfrac{x+2y+3z}{5+8+9}\) = \(\dfrac{x+2y+3z}{22}\)

\(\dfrac{x}{5}\)= \(\dfrac{y}{4}\) = \(\dfrac{z}{3}\) = \(\dfrac{x-2y+3z}{5-8+9}\) = \(\dfrac{x-2y+3z}{6}\)

=> \(\dfrac{x+2y+3z}{22}\) = \(\dfrac{x-2y+3z}{6}\)

=> \(\dfrac{x+2y+3z}{x-2y+3z}\) = \(\dfrac{22}{6}\) =\(\dfrac{11}{3}\)