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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).
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)
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
\(A=\dfrac{x}{xy+x+1}+\dfrac{xy}{x.yz+xy+x}+\dfrac{xy.z}{xy.xz+xy.z+xy}\)
\(=\dfrac{x}{xy+x+1}+\dfrac{xy}{1+xy+x}+\dfrac{1}{x+1+xy}\)
\(=\dfrac{x+xy+1}{xy+x+1}=1\)
Đặt \(A=\dfrac{2014x}{xy+2014x+2014}+\dfrac{y}{yz+y+2014}+\dfrac{z}{xz+z+1}\)
\(A=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)
\(A=\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)
\(A=\dfrac{xz}{xz+z+1}+\dfrac{1}{xz+z+1}+\dfrac{z}{xz+z+1}\)
\(A=\dfrac{xz+z+1}{xz+z+1}=1\)
\(\Rightarrowđpcm\)
Ta có : \(A=\dfrac{2014x}{xy+2014x+2014}+\dfrac{y}{yz+y+2014}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{xyz.x}{xy+xyz.x+xyz}+\dfrac{x.y}{x.yz+xy+xyz.x}+\dfrac{xy.z}{xz.xy+xy.z+xy}\)
\(=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{xy}{xyz+x^2yz+xy}+\dfrac{xyz}{x^2yz+xyz+xy}\)
\(=\dfrac{x^2yz+xyz+xy}{x^2yz+xyz+xy}=1\) (const)
Vậy A không phụ thuộc vào các biến x,y,z
\(\dfrac{2006x}{xy+2006x+2006}+\dfrac{y}{yz+y+2006}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(xz+z+1\right)}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{xz}{xz+z+1}+\dfrac{1}{xz+z+1}+\dfrac{z}{xz+z+1}=\dfrac{xz+z+1}{xz+z+1}=1\)
Ta có: \(\dfrac{2006x}{xy+2006x+2006}+\dfrac{y}{yz+y+2006}+\dfrac{z}{xz+z+1}=1\)
\(\Leftrightarrow\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+x+1}\)
\(\Leftrightarrow\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+x+1}\)
\(\Leftrightarrow\dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+x+1}\)
\(\Leftrightarrow\dfrac{xz+1+z}{1+xz+z}=1\left(đpcm\right)\)
_Chúc bạn học tốt_