Ta có:

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

                                              \(=\frac{xyz}{x.\left(y+1+yz\right)}+\frac{y}{yz+y+1}+\frac{1}{1+yz+y}\)

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

                                               \(=\frac{yz+y+1}{yz+y+1}=1\left(đpcm\right)\)

bạn cho mình biết sau dấu + bị che khuất là số nào được k?

= 1 nhé

thay x.y.z zô biểu thức đi . rùi đặt nhân tử chung rùi tự làm , đến đó mà k làm dc nữa  thì die đi

\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)

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

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

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

=>đpcm

2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1

= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1

= xz/1+xz+z + 1/z+1+xz + z/xz+z+1

= xz+1+x/1+xz+x = 1 (đpcm)

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 \(x=\frac{a}{b};y=\frac{b}{c};z=\frac{c}{a}\)

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

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

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

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

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