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18 tháng 1 2019

\(\dfrac{x}{xyz+xy+x+1}+\dfrac{y}{yzt+yz+y+1}+\dfrac{z}{xzt+zt+z+1}+\dfrac{t}{xyt+tx+t+1}\)

= \(\dfrac{x}{xyz+xy+x+1}+\dfrac{xy}{xyzt+xyz+xy+x}+\dfrac{xyz}{x^2yzt+xyzt+xyz+xy}+\dfrac{xyzt}{x^{2^{ }}y^2zt+x^2yzt+xyzt+xyz}\)

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

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

= 1

14 tháng 1 2019

Thay xyzt = 1 vào P, có:

P= \(\frac{x}{xyz+xy+x+xyzt\ }\) + \(\frac{y}{yzt+yz+y+1}+\frac{z}{xzt+zt+z+xyzt}+\frac{t}{xyt+tx+t+1}\)

\(P=\frac{x}{x.\left(yz+y+1+yzt\right)}+\frac{y}{yzt+yz+y+1}+\frac{z}{z.\left(xt+t+1+xyt\right)}+\frac{t}{xyt+tx+t+1}\)

\(P=\frac{1\ +y}{yz+y+yzt+1}\) \(+\frac{1+t}{xyt+tx+t+1}\)

\(P=\frac{1+y}{yz+y+yzt+xyzt\ }+\frac{1+t}{xyt+tx+t+1}\)

\(P=\frac{1+y}{y.z.\left(xyt+tx+t+1\right)}+\frac{yz+tyz}{yz.\left(xyt+tx+t+1\right)}\)

\(P=\frac{1+y+yz+tyz}{yz.\left(xyt+tx+t+1\right)}=\frac{1+y+yz+tyz}{xyzt.\left(1+y+yz+tyz\right)}=\frac{1}{xyzt}=1\)

KL: P = 1 tại xyzt=1

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

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 

23 tháng 12 2018

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

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

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

31 tháng 7 2018

abcd=1 đâu ra zậy

\(S=\left(xy+yz+zx\right)\cdot\frac{xy+yz+zx}{xyz}-\frac{xyz\left(x^2y^2+y^2z^2+z^2x^2\right)}{x^2y^2z^2}\)

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

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

\(=\frac{2xyz\left(x+y+z\right)}{xyz}=2\left(x+y+z\right)\)

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~