Chứng minh P=(x+y)/(z+t)+(y+z)/(t+x)+(z+t)/(x+y)+(t+x)/(y+z) có giá trị nguyên

Sửa đề bài : P = \(\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\)

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

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

=> \(\dfrac{y+z+t}{x}+1=\dfrac{z+t+x}{y}+1=\dfrac{t+x+y}{z}+1=\dfrac{x+y+z}{t}+1\)=> \(\dfrac{y+z+t+x}{x}=\dfrac{z+t+x+y}{y}=\dfrac{t+x+y+z}{z}=\dfrac{x+y+z+t}{t}\)TH1: x + y + z + t # 0

=> x = y = z = t

Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)

P = \(\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\)

P = 1 + 1 + 1 + 1 = 4

TH2 : x + y + z + t = 0

=> x + y = -(z + t)

y + z = -(t + x)

z + t = -(x + y)

t + x = -(y + z)

Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)

P = \(\dfrac{-\left(z+t\right)}{z+t}=\dfrac{-\left(t+x\right)}{t+x}=\dfrac{-\left(x+y\right)}{x+y}=\dfrac{-\left(y+z\right)}{y+z}\)

P = (-1) + (-1) + (-1) + (-1)

P = -4

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