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 ba số x,y,z biết rằng \(2x=\dfrac{y}{3}=\dfrac{z}{5}\)và \(x+y-\dfrac{z}{2}=-20\)
Tìm x,y,z biết:\(\dfrac{x}{y+z-5}=\dfrac{y}{x+z+3}=\dfrac{z}{x+y+2}=\dfrac{1}{2}\)
tìm x,y,z biết \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
a) Chứng minh rằng nếu 2(x+y) = 5(y+z) = 3(z+x)
Thì \(\dfrac{x-y}{4}=\dfrac{y-z}{5}\)
b) Cho \(x^2=yz\) . Chứng minh rằng \(\dfrac{x^2+y^2}{y^2+z^2}=\dfrac{x}{z}\)
cho x,y,z khac 0 va\(\dfrac{x+3y-z}{z}\)= \(\dfrac{y+3z-x}{x}\)=\(\dfrac{z+3x-y}{y}\)
Tính P = \(\left(\dfrac{x}{y}+3\right)\)\(\left(\dfrac{y}{z}+3\right)\)\(\left(\dfrac{z}{x}+3\right)\)
Tìm x, y, z biết
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+3}=\dfrac{z}{y+x-4}=x+y+z\)
tìm x,y,z
x+y+z=\(\dfrac{x}{y+z-2}=\dfrac{y}{z+x-3}=\dfrac{z}{x+y+5}\)
Tìm x,y,z biết: \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\)và x+y-z=10