cho;x/y+z+t=y/z+t+x=z/t+x+y=t/x+y+z cm.Tính giá trị biểu thức: P=(x+y/z+t)+(y+z/t+x)+(z+t/y+x)+(t+x/y+z)

Áp dụng tính chất dãy các tỉ số bằng nhau, ta có:

`x/(y + z + t) = y/(z + t + x) = z/(t + x + y) = t/(x + y + z)`

`= (x + y + z + t)/(y + z + t + z + t + x + t + x + y + x + y + z)`

`= (x + y + z + t)/(3x + 3y + 3z + 3t)`

`= (x + y + z + t)/(3(x + y + z + t))`

`= 1/3`

`=>` \(\left\{\begin{matrix}\dfrac{x}{y+z+t}=\dfrac{1}{3}\\\dfrac{y}{z + t + x}=\dfrac{1}{3}\\\dfrac{z}{t + x + y}=\dfrac{1}{3}\\\dfrac{t}{x + y + z}=\dfrac{1}{3}\end{matrix}\right.\)

`=>` \(\left\{\begin{matrix}3x=y+z+t\\3y=z+t+x\\3z=t+x+y\\3t=x+y+z\end{matrix}\right.\)

`=>` \(\left\{\begin{matrix}4x=y+z+t+x\\4y=z+t+x+y\\4z=t+x+y+z\\4t=x+y+z+t\end{matrix}\right.\)

`=> 4x = 4y = 4z = 4t`
`=> x = y = z = t`
`=>` \(\left\{\begin{matrix}\dfrac{x+y}{z+t}=1\\\dfrac{y+z}{ t + x}=1\\\dfrac{z+t}{y+x}=1\\\dfrac{t+x}{y + z}=1\end{matrix}\right.\)

`=> (x + y)/(z + t) + (y + z)/(t + x) + (z + t)/(y + x) + (t + x)/(y + z) = 1 + 1 + 1 + 1 = 4`

`=> P = 4`

Vậy `P = 4`