Question

[Long Lê]

Cho xyz=1

Tính tổng \(T=\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\)

Giúp nhanh nha mọi người

Answer 1

[Mysterious Person]

ta có : \(T=\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{x}{xyz+xy+x}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{1}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{y+1}{yz+y+1}+\dfrac{z}{xz+z+1}=\dfrac{xyz+y}{xyz+yz+y}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{xz+1}{xz+z+1}+\dfrac{z}{xz+z+1}=\dfrac{xz+z+1}{xz+z+1}=1\)