Question

[Nguyễn Minh Hoàng]

Tìm x, y, z :

\(\frac{3x-2y}{5}=\frac{2z-5x}{3}=\frac{5y-3z}{2}\) và x + y + z = 100

Answer 1

[.]

\(\frac{15x-10y}{5^2}\)=\(\frac{6z-15x}{3^2}\)=\(\frac{10y-6z}{2^2}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{15x-10y}{5^2}\)=\(\frac{6z-15x}{3^2}\)=\(\frac{10y-6z}{2^2}\)=\(\frac{15x-10y+6z-15x+10y-6z}{5^2+3^2+2^2}\)=0

Suy ra 3x=2y                          \(\frac{x}{2}\)=\(\frac{y}{3}\)

           2z=5x               Suy ra \(\frac{z}{5}\)=\(\frac{x}{2}\)

            5y=3z

Suy ra \(\frac{x}{2}\)=\(\frac{y}{3}\)=\(\frac{z}{5}\)

áp dụng t/c dãy tỉ số bằng nhau

\(\frac{x}{2}\)=\(\frac{y}{3}\)=\(\frac{z}{5}\)=\(\frac{x+y+z}{2+3+5}\)=\(\frac{100}{10}\)=10

 x/2=10 suy ra x=20

y/3=10 suy ra y=30

z/3=10 suy ra z=50

k cho mình nha <3