Question

cho xyz=2006 chứng minh rằng

\(\frac{2006x}{xy+2006x+2006}\) + \(\frac{y}{yz+y+2006}\) + \(\frac{z}{zx+z+1}\) =1

Answer 1

Thay 2006=xyz

Ta có : 

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

\(=>\frac{x^2yz}{xy\left(zx+z+1\right)}+\frac{y}{y\left(zx+z+1\right)}+\frac{z}{zx+x+1}\)

=> \(\frac{xz}{zx+z+1}+\frac{1}{zx+z+1}+\frac{z}{zx+x+1}\)= 1(điều phải chứng minh)

 

Answer 2

Ta có: \(A=\frac{2006x}{xy+2006x+2006}+\frac{y}{yz+y+2006}\) \(+\frac{z}{zx+z+1}\)

\(=\frac{2006xz}{xyz+2006zx+2006z}+\frac{y}{yz+y+xyz}\) \(+\frac{z}{zx+z+1}\)

\(=\frac{2016xz}{2016\left(1+zx+z\right)}+\frac{y}{y\left(z+1+xz\right)}\) \(+\frac{z}{zx+z+1}\)

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

=> đpcm