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18 tháng 3 2019

Theo bài ra, ta có:

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

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

\(=\frac{x}{xy+x+1}+\frac{xy}{xyz+xy+x}+\frac{z}{z\left(x+1+xy\right)}\)

\(=\frac{x}{xy+x+1}+\frac{xy}{xy+x+1}+\frac{1}{xy+x+1}\)

\(=\frac{x+xy+1}{xy+x+1}\)

\(=1\)

Vậy P = 1

18 tháng 3 2019

Ta có: P = \(\dfrac{x}{xy+x+1}\)+\(\dfrac{y}{yz+y+1}\)+\(\dfrac{z}{xz+z+1}\)

=\(\dfrac{x}{xy+x+1}\)+\(\dfrac{xy}{xyz+xy+x}\)+\(\dfrac{xyz}{x^2yz+xyz+xy}\)

=\(\dfrac{x}{xy+x+1}\)+\(\dfrac{xy}{xy+x+1}\)+\(\dfrac{1}{xy+x+1}\)(vì xyz=1)

=\(\dfrac{x+xy+1}{xy+x+1}\)

=1

Vậy P = 1

23 tháng 12 2018

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

\(B=\frac{x}{xy+x+xyz}+\frac{y}{yz+y+1}+\frac{yz}{xyz+yz+y}\)

\(B=\frac{1}{y+1+yz}+\frac{y}{y+1+yz}+\frac{yz}{y+1+yz}=1\)

10 tháng 6 2015

Từ xyz=1

=>\(A=\frac{x}{-xy+x+1}-\frac{y}{yz-y+1}+\frac{z}{xz+z-1}\)

=\(\frac{xz}{-xyz+xz+z}-\frac{xyz}{xyz^2-xyz+xz}+\frac{z}{xz+z-1}\)

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

3 tháng 12 2018

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

\(M=\frac{xyz}{x\left(1+y+yz\right)}+\frac{1}{1+y+yz}+\frac{y}{y+yz+xyz}\)

\(M=\frac{yz}{1+y+yz}+\frac{1}{1+y+yz}+\frac{y}{y+yz+1}\)

\(M=\frac{yz+y+1}{1+y+yz}\)

Tham khảo nhé~

2 tháng 1 2016

ta có x/xy+x+1 +y/yz+y+1 +z/xz+z+1

=xz/xyz+xz+z +xyz/xyz^2+xyz+xz +z/xz+z+1

=xz/1+xz+z +1/z+1+xz +z/ xz+z+1

=xz+z+1 /xz+z+1 =1

15 tháng 12 2016

T/c:xyz=1

=>x=1;y=1;z=1

=>T=1/1+1+1   +1/1+1+1   +1/1+1+1

=>T=1/3  +1/3  +1/3

=>T=1

15 tháng 12 2016

Ta co : x.y.z=1

Hay : x=1 ; y=1 va z=1

\(\Rightarrow T=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)

\(=\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}\)

\(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)

\(\Rightarrow\)T=1 

24 tháng 1 2017

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

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

\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)

\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)

\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)

\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)

\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)

\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)

\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)

\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)

Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)

7 tháng 10 2016

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)

\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2=xy^2-x^2z-xy^3z+x^2yz^2\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)

\(\Rightarrow xy\left(x-y\right)-xyz\left(x-y\right)\left(x+y+z\right)+z\left(x-y\right)\left(x+y\right)=0\)

\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xz+yz\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)

Mà \(x\ne y\) nên \(xy+xz+yz-xyz\left(x+y+z\right)=0\)

\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

Đpcm

7 tháng 10 2016

Từ gt ta có : (x2 - yz)y(1 - yz) = (y2 - xz)x(1 - yz)

=> 0 = VT - VP = (x2y - x3yz - y2z - xy2z2) - (xy2 - xy3z  - x2z - x2yz2) = xy(x - y) - xyz(x2 - y2) + z(x2 - y2) + xyz2(y - x)

= (x - y)[xy - xyz(x + y) + z(x + y) - xyz2] = (x - y)(xy + yz + xz - xyz(x + y + z)]

\(x\ne y\Rightarrow x-y\ne0\) nên xy + yz + xz - xyz(x + y + z) = 0 => xy + yz + xz = xyz(x + y + z)

Bạn ko hiểu chỗ nào thì hỏi mình nhé!