Question

cho \(\dfrac{x}{y+z+t}\) = \(\dfrac{y}{z+t+x}\) = \(\dfrac{z}{t+x+y}\) = \(\dfrac{t}{x+y+z}\)

CMR : P = \(\dfrac{x+y}{z+t}\) + \(\dfrac{y+z}{t+x}\) + \(\dfrac{z+t}{x+y}\) + \(\dfrac{t+x}{y+z}\) \(\in\) Z

Answer 1

Ta có:

\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)

\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)

\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{y+x+x}=\dfrac{x+y+z+t}{y+x+z}\)

. Xét TH1: \(x+y+z+t=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\end{matrix}\right.\)

. Xét TH2: \(x+y+z+t\ne0\)

\(\Rightarrow x=y=z=t\)

\(\Rightarrow A=1\)

\(\Rightarrow\left\{{}\begin{matrix}A=1\\A=-1\end{matrix}\right.\)

Answer 2

P =4