Question

cho x/2 = y/3 ; y/4 = z/5  tính  x+y+z / 2x+3y+4z

Answer 1

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{2x}{16}=\dfrac{3y}{36}=\dfrac{4z}{60}=\dfrac{x+y+z}{35}=\dfrac{2x+3y+4z}{112}\\ \Rightarrow\dfrac{x+y+z}{2x+3y+4z}=\dfrac{35}{112}=\dfrac{5}{16}\)

Answer 2

          \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x}{8}=\dfrac{y}{12};\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

*  \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y+z}{8+12+15}=\dfrac{x+y+z}{45}\)                                       (1)

* \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{2x}{16}=\dfrac{3y}{36}=\dfrac{4z}{60}=\dfrac{2x+3y+4z}{16+36+60}=\dfrac{2x+3y+4z}{112}\)   (2)

(1)(2)=> \(\dfrac{x+y+z}{45}=\dfrac{2x+3y+4z}{112}=\dfrac{x+y+z}{2x+3y+4z}=\dfrac{45}{112}\)

         => A = \(\dfrac{45}{112}\)