Question

Tìm x,y, z

2x/3 = 3y/4=4z/5 và x+y+z=49

 

Answer 1

Ta có : \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=\)\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)

Nên : \(\frac{2x}{3}=12\Rightarrow2x=36\Rightarrow x=18\)

         \(\frac{3y}{4}=12\Rightarrow3y=48\Rightarrow y=16\)

          \(\frac{4z}{5}=12\Rightarrow4z=60\Rightarrow z=15\)

Vậy x = 18 ; y = 16 ; z = 15