Các câu hỏi tương tự
Tim x , y , z biet:
\(\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{z}{x+y+3}=x+y+z\)
Tim ba so x, y, z biet \(\frac{y+z+1}{x}=\frac{x+ z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
Tim x,y,z biet rang: \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
tim x,y,z biet
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z\)
Tim x , y , z biet:
\(\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{z}{x+y-3}=x+y+z\)
Cach lam ho minh voi
Tim x;y;z biet:
\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)
Tim x,y,z biet:
a, xy=z ; yz=4x ; zx=9x
b, \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
Tim x,y,z biet:
a, \(xy=z;yz=4x;zx=9y\)
b, \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
tim x,y,z biet \(\frac{3.X-5.Y}{2}=\frac{5.Y-3.Z}{3}=\frac{3.B}{4};X+Y+Z+17\)=17