Question

Tìm x,y,z

x:y:z=3:4:5 và x-2y+32=60

Answer 1

Ta có : x:y:z = 3:4:5

 \(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{2y}{8}=\frac{3z}{15}=\frac{z-2y+3z}{3-8+15}=\frac{60}{10}=6\)

Nên : \(\frac{x}{3}=6\Rightarrow x=18\)

         \(\frac{y}{4}=6\Rightarrow y=24\)

          \(\frac{z}{5}=6\Rightarrow z=30\)

Vậy x = 18 ; y = 24 ; z = 30

Answer 2

x:y:z=3:4:5 => x/3 = y/4 = z/5 => x/3 = 2y/8 = 3z/15

Theo TCDTSBN ta có:

x/3 = 2y/8 = 3z/15 = x-2y+3z/3-8+15 = 60/10 = 6

=> x = 18, y = 24, z = 30