\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5},xyz=810\)

\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

tìm x,y,z(nếu có

Tìm x,y,z biết

a) \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

b) \(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{z}{x+y-3}=x+y=z\)

Tìm x,y,z biết

a) \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

b) \(\frac{x}{y+z+1}=\frac{y}{x+z=2}=\frac{z}{x+y-3}=x+y=z\)

Tìm x,y,z biết

a) \(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{z}{x+y-3}=x+y+z\)

b) \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

a) tìm x,y,z biết rằng \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

b) tìm x biết \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\) 

\(a,\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

\(b,\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

\(c,\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)

Tìm các số x, y, z biết rằng:

a) \(\frac{y+z+1}{x}=\frac{x+y+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\);

b) \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\);

c) \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)

1/

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

b/ \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

c/ \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

d/ \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)

bài 1 :tìm x, y, z:

\(\frac{x}{x+y+z}=\frac{y}{x+z+2}=\frac{z}{x+y-3}=x+y+z\left(1\right)\)

bài 2:tìm x, y:

a)\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)

b)\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)