Violympic toán 7
Các câu hỏi tương tự
Tìm x;y;z biết
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
tìm x,y,z biết \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
tìm x,y,z biết \(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y+2}=x+y+z\)
Biết x+y+z = 2017 và \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=\dfrac{1}{672}\)
Tính tổng A = \(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\)
Tìm x,y,z biết :
\(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\) (\(x,y,z\ne0\))
Tìm x,y,z biết
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+z-2}=x+y+z\)
tìm x,y,z khác 0 biết \(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)
Tìm x,y,z biết :
\(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y+2}=x+y+z\) (x,y,z \(\ne\) 0)
Tìm x,y,z biết:\(\dfrac{x}{y+z-5}=\dfrac{y}{x+z+3}=\dfrac{z}{x+y+2}=\dfrac{1}{2}\)