tìm x,y biết \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
Tìm cặp số x;y biết : \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
Tìm các cặp số ( x , y) biết :
\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
Tìm cặp số (x,y)
\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
Tìm x, y:
\(\dfrac{1+7y}{4x}=\dfrac{1+5y}{5x}=\dfrac{1+3y}{12}\)
Giúp tớ với....
\(Tìmx,y,z\\ \dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
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, \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}va2x^2+2y^2-z^2=1\)
\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
tìm x,y,z
a)\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
b)\((x-1)=2\left(y-2\right)=3\left(z-3\right)\)và \(2x+3y-z=50\)
c) \(\dfrac{3x-2y}{37}=\dfrac{5y-3z}{15}=\dfrac{2z-5x}{2}\) và \(10-3y-2z=-4\)
d) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)và \(2x+3y-z=50\)
e) \(ab=\dfrac{1}{2};bc=\dfrac{2}{3};ac=\dfrac{3}{4}\)
f)\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\) và \(x.y.z=22400\)