Cho xyz =1. Tính : \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)
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S
31 tháng 7 2018
abcd=1 đâu ra zậy
\(S=\left(xy+yz+zx\right)\cdot\frac{xy+yz+zx}{xyz}-\frac{xyz\left(x^2y^2+y^2z^2+z^2x^2\right)}{x^2y^2z^2}\)
\(=\frac{\left(xy+yz+zx\right)^2}{xyz}-\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\)
\(=\frac{x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)-x^2y^2-y^2z^2-z^2x^2}{xyz}\)
\(=\frac{2xyz\left(x+y+z\right)}{xyz}=2\left(x+y+z\right)\)
PT
3
5 tháng 12 2014
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
LV
1 tháng 10 2017
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
\(=\frac{x+xy+1}{xy+x+1}\)
\(=1\)
12 tháng 11 2018
\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
XO
11 tháng 3 2020
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
Lời giải:
Ta có:
\(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)
\(=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+zxy+zx.xy}\)
\(=\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xy}{xy+1+x}=\frac{1+x+xy}{1+x+xy}=1\) (thay $xyz=1$)
$\Rightarrow $ đpcm