tìm x y 1+3y/12=1+6y/5x=1+9y/4x
tìm x , y , z biết
a,
\(\frac{x+y}{x}=\frac{y}{x+z}=\frac{z}{x+y}=x+y+z\)
b,
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)
c,
\(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}\)
d,
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
tìm x thỏa mãn: \(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
1+3y/12=1+6y/16=1+9y/4x
Tìm x thỏa mãn: \(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
Tìm x thỏa mãn:
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
2) Tìm x, y biết \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)(với x, y khác 0)
Tìm x,y,z biết:
a) \(\dfrac{x}{3}=\dfrac{z}{8}\); -6y = 7z và 2x - 9y = 2
b) \(\left|4-2x\right|+\left|x-2\right|=3-x\)
c) (5x-3)2013 = (5x-3)2015
a,\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{+6x}\)
b, \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
c,\(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\left(x,y,zkhac0\right)\)
d, \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}va2x^2+2y^2-z^2=1\)
