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25 tháng 12 2016

Xét: \(x+y+z=xyz\Leftrightarrow\frac{x+y+z}{xyz}=1\)

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

Mặt khác:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}\)<=>\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\left(\sqrt{3}\right)^2\)

<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=3\)

<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=3\)

<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2.1=3\)

<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2=3\)

<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1\)

7 tháng 4 2017

ủng hộ mk nha mọi người

16 tháng 2 2018

\(x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}\)\(\Rightarrow\hept{\begin{cases}x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{xy}\\y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}\\z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}\end{cases}}\)

\(\Rightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(xyz\right)^2}\)

\(\Leftrightarrow\frac{1}{\left(xyz\right)^2}=1\Rightarrow xyz=\pm1\)(đpcm)

1 tháng 1 2016

3x²y²z² = x³y³ y³z³ z³x³ 
(3x²y²z²) / (x³y³ y³z³ z³x³) = 1
3.[(x²y²z²) / (x³y³ y³z³ z³x³)] = 1
(x²y²z²) / (x³y³ y³z³ z³x³) = 1/3
(x²y²z²) / (x³y³) (x²y²z²) / (y³z³) (x²y²z²) / (z³x³) = 1/3
z²/(xy) x/(yz) y²/(zx) = 1/3
Vậy x²/(yz) y²/(xz) z²/(xy) = 1/3

12 tháng 12 2016

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

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

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

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

=>đpcm

12 tháng 12 2016

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)

18 tháng 11 2021

\(x^2=y.z\Rightarrow x^3=x.y.z\\ y^2=x.z\Rightarrow y^3=x.y.z\\ z^2=x.y\Rightarrow z^3=x.y.z\\ \Rightarrow x^3=y^3=z^3\\ \Rightarrow x=y=z\)

17 tháng 4 2022

Ta có \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\)

\(\Leftrightarrow\dfrac{\left(yz\right)^2+\left(xz\right)^2+\left(xy\right)^2+2xyz}{\left(xyz\right)^2}=1\)

<=> (xy)2 + (yz)2 + (zx)2 + 2xyz = (xyz)2 

<=> (xy)2 + (yz)2 + (xz)2 + 2xyz(x + y + z) = (xyz)2 

<=> (xy + yz + zx)2 = (xyz)2 

<=> \(\left[{}\begin{matrix}xy+yz+zx=xyz\\xy+yz+zx=-xyz\end{matrix}\right.\)

+) Khi xy + yz + zx = -xyz 

=> \(\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-1< 0\left(\text{loại}\right)\)

=> xy + yz + zx = xyz 

<=> \(xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=xyz\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1\right)=0\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

<=> \(\dfrac{x+y}{xy}=\dfrac{-\left(x+y\right)}{\left(x+y+z\right)z}\)

<=> \(\left(x+y\right)\left(\dfrac{1}{xz+yz+z^2}+\dfrac{1}{xy}\right)=0\)

<=> \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(zx+yz+z^2\right)xy}=0\)

<=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

Khi x = -y => y = 1 => P = 1

Tương tự y = -z ; z = -x được P = 1

Vậy P = 1 

17 tháng 4 2022

tks b nha

 

19 tháng 9 2019

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=3\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=3\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}=3\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2xyz}{xyz}=3\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2=3\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1\)