\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{y+z+t}{3}\\y=\dfrac{z+t+x}{3}\\z=\dfrac{t+x+y}{3}\\t=\dfrac{x+y+z}{3}\end{matrix}\right.\)\(\Rightarrow x=y=z=t\)
cho 4 số x,y,z thỏa mãn:\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{y+t+x}\dfrac{t}{y+x+z}\)
Hãy tính giá trị của biểu thức:\(A=\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
Cho biểu thức: \(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{\:t+x}{y+z}\) . Tìm giá trị của P biết rằng: \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\) .
Cho \(\dfrac{x}{y+z+t}\)= \(\dfrac{y}{z+t+x}\)= \(\dfrac{z}{t+x+y}\)= \(\dfrac{t}{x+y+z}\)( giả thuyết các tỉ số đều có nghĩa )
Câu 1 : Biết\(\dfrac{x}{t}=\dfrac{5}{6};\dfrac{y}{z}=\dfrac{1}{5};\dfrac{z}{x}=\dfrac{7}{3}\) ( x; y; z; t khác 0 ). Hãy tìm tỉ số \(\dfrac{t}{y}\)
A. \(\dfrac{t}{y}=\dfrac{14}{25}\) B. \(\dfrac{t}{y}=\dfrac{7}{8}\) C. \(\dfrac{t}{y}=\dfrac{18}{7}\) D. \(\dfrac{t}{y}=\dfrac{6}{7}\)
Cho x,y,z,t ∈ N. CM :
M = \(\dfrac{x}{x+y+z}=\dfrac{y}{x+y+t}=\dfrac{z}{z+t+x}\) ∉ N
Tìm x, y, z
\(\dfrac{x+y+2017}{z}=\dfrac{y+z-2018}{x}=\dfrac{z+x+1}{y}=\dfrac{2}{x+y+z}\)
Tìm x; y; z
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Tìm x, y, z biết:
\(\dfrac{x+y-3}{z}=\dfrac{x+z+2}{y}=\dfrac{y+z+1}{x}=\dfrac{1}{x+y+z}\)
1. Tìm x, y, z:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
