Question

Bài 1: Cho ba số x,y,z thỏa mãn x.y.z = 2017

Tính tổng D = \(\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{zx+z+1}\)

Thông cảm vì mk đăng ko đúng dạng bài nhưng mong các bn giúp mk vs ak. Mk cảm ơn

Answer 1

\(D=\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{zx+z+1}\)

\(D=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
Vì \(xyz=2017\)
\(D=\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz+1+z}{1+xz+z}=1\)
Vậy D = 1