Ta có: xyz=2006
Đặt tổng (đề) trên là A ( phân số thứ nhất tử là 2006x nhé)
=> \(A=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=\frac{xz+1+z}{xz+z+1}=1\)
=> A = 1 (đpcm).
cho xyz=2006 . chung minh rang \(\frac{2006x}{xy+2006x+2006}+\frac{y}{yz+y+2006}+\frac{z}{xz+z+1}=1\)
Cho xyz = 2006 Chứng minh \(\frac{2006x}{xy+2006x+2006}\)+ \(\frac{y}{yz+2006x+2006}\)+\(\frac{z}{xz+z+1}\)= 1
cho xyz=2006 chứng minh rằng
\(\frac{2006x}{xy+2006x+2006}\) + \(\frac{y}{yz+y+2006}\) + \(\frac{z}{zx+z+1}\) =1
cho xyz=2006.chứng minh rằng [2006x/(xy+2006x+2006)]+[y/(yz+y+2006)]+[z/(xz+z+1)]=1 làm nhanh giúp tôi nha
Cho xyz = 1. Tính B = \(\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)
Cho xyz = 2017
CMR : \(\frac{2017x}{xy+2017x+2017}+\frac{y}{yz+y+2017}+\frac{z}{xz+z+1}=1\)
cho xyz=2018
tính \(\frac{2018x}{xy+2018x+2018}+\frac{y}{yz+y+2018}+\frac{z}{xz+z+1}\)
Cho xyz=2011
Chứng minh rằng \(\frac{2011x}{xy+2011x+2011}+\frac{y}{yz+y+2011}+\frac{z}{xz+z+1}=1\)
Cho xyz=1
Tính A=\(\frac{x}{-xy+x+1}-\frac{y}{yz-y+1}+\frac{z}{xz+z-1}\)
