Ta có : \(A=\frac{2019}{x+xy+1}+\frac{2019}{y+yz+1}+\frac{2019}{z+zx+1}=2019\left(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+xyz+z}+\frac{xz}{xyz+xyz^2+xz}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+z+1}+\frac{xz}{1+z+xz}+\frac{1}{z+zx+1}\right)\)(vì xyz = 1)

\(=2019\left(\frac{z+xz+1}{xz+z+1}\right)=2019\)

Vậy A = 2019

Do \(xyz=1\)nên:

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

\(=\frac{1}{xy+x+1}+\frac{x}{xyz+xy+z}+\frac{xy}{x^2yz+xyz+xy}\)

\(=\frac{1}{xy+x+1}+\frac{x}{1+xy+x}+\frac{xy}{x+1+y}=1\)

=> ĐPCM

\(xyz=1\) nên tồn tại \(x=\frac{a}{b};y=\frac{b}{c};z=\frac{c}{a}\)

\(\frac{1}{xy+x+1}+\frac{1}{yz+y+1}+\frac{1}{zx+z+1}\)

\(=\frac{1}{\frac{a}{b}\cdot\frac{b}{c}+\frac{a}{b}+1}+\frac{1}{\frac{b}{c}\cdot\frac{c}{a}+\frac{b}{c}+1}+\frac{1}{\frac{c}{a}\cdot\frac{a}{b}+\frac{c}{a}+1}\)

\(=\frac{1}{\frac{a}{c}+\frac{a}{b}+1}+\frac{1}{\frac{b}{a}+\frac{b}{c}+1}+\frac{1}{\frac{c}{b}+\frac{c}{a}+1}\)

\(=\frac{bc}{ab+ac+cb}+\frac{ac}{bc+ab+ac}+\frac{ab}{ac+bc+ab}\)

\(=\frac{ab+bc+ca}{ab+bc+ca}=1\)

x+y+z hay là xyz hả bạn

x*y*z =2018 nha

TA CÓ \(\frac{x}{xy+x+1}\)+\(\frac{y}{yz+y+1}\)+\(\frac{z}{xz+z+1}\)

         =\(\frac{x}{xy+x+1}\)+\(\frac{xy}{xyz+xy+x}\)+\(\frac{xyz}{x^2yz+xyz+xy}\)

         =\(\frac{x}{xy+x+1}\)+\(\frac{xy}{xy+x+1}\)+\(\frac{1}{xy+x+1}\)(vì xyz=1)

        =\(\frac{x+xy+1}{xy+x+1}\)

        = 1

ta có :

\(\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{xyz+yz+y}\)

\(\frac{xyz}{xy+x+xyz}+\frac{y}{yz+y+1}+\frac{xyz}{1+yz+y}\)

\(\frac{yz+y+xyz}{y+1+yz}\)

\(\frac{yz+y+1}{yz+y+1}\)

=1

luffy123 làm đúng mà sao vẫn có đứa bảo sai kìa

Ta có: \(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)

\(A=\frac{xz}{xyz+xz+z}+\frac{xyz}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}\)

\(A=\frac{xz}{1+xz+z}+\frac{xyz}{z+1+xz}+\frac{z}{xz+z+1}\)

\(A=\frac{xyz+xz+1}{xyz+xz+1}\)

\(A=1\)

Vậy \(A=1\)

GOOD