Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{-3}=\dfrac{y}{12}=\dfrac{x+y}{-3+12}=\dfrac{18}{9}=2\)
\(\dfrac{x}{-3}=2\Rightarrow x=-6\\ \dfrac{y}{12}=2\Rightarrow y=24\)
Tìm x,y,z biết:
a) \(\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=9\)
b) \(\dfrac{x-3}{12}=\dfrac{-3}{3-x}\)
c) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và \(x-y-z=-49\)
\(\dfrac{x}{y+z-3}=\dfrac{y}{x+z}=\dfrac{z}{x+y+3}\left(x+y+z=\dfrac{1}{12}\right)\)
\(\dfrac{x}{5}\) = \(\dfrac{y}{4}\) =\(\dfrac{z}{3}\) và x + y - z =18
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(\text{x+y+z}=18\)
a,\(\dfrac{x}{3}=\dfrac{y}{4}\)va x+y= 28 b, \(\dfrac{x}{9}=\dfrac{4}{x}\)
Can co nguoi giai nhanh ah ! em cam on
12) Tìm x, y ϵ Z, sao cho:
a) \(\dfrac{x}{2}\) - \(\dfrac{1}{y}\)= \(\dfrac{1}{3}\)
b) \(\dfrac{4}{x}\) + \(\dfrac{y}{2}\) = \(\dfrac{-1}{4}\)
tìm x, y ,z biết
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\) và x+y+z = 18
Tìm x,y,z biết \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)và x+y+z=18
Tìm hai số x, y biết \(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)và x+y = − 32
A. =− 20; =− 12
B. =− 12; = 20
C. =− 12; =− 20
D. = 12; =− 20
