Question

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)và x+y+z=49

Answer 1

Answer 2

Ta có \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

=> \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)=\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Theo tính chất của dãy tỉ số bằng nhau ta được:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)=\(\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}\)=12

=> x = \(\dfrac{3}{2}.12=18\)

y=12. 4/3=16

z= 12. 5/4= 15

Answer 3

Ta có:

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

\(=>\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:

\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\)\(\dfrac{12x+12y+12z}{18+16+15}=\dfrac{12\left(x+y+z\right)}{18+16+15}=\dfrac{12.49}{49}=12\)

\(\dfrac{12x}{18}=12=>12x=216\)

\(\dfrac{12y}{16}=12=>12y=192\)

\(\dfrac{12z}{15}=12=>12z=180\)

=> \(x=18;y=16;z=15\)