Question

x,y,z three variable and y+z-x is constantand (x+y-z)(x+z-x) very as yz then show that (x+y+z) very as yz

Answer 1

Answer:

Given:- xyz=1

To prove:- (1+x+y −1 ) ^−1

+(1+y+z −1 ) ^−1

+(1+z+x −1 ) −1 =1

Proof:-

Taking L.H.S.-

(1+x+y ^−1 ) −1

+(1+y+z −1 )^ −1

+(1+z+x ^−1 ) ^−1

=(1+x+xz) ^−1 +(1+y+xy) ^−1 +(1+z+yz) ^−1

=(1+x+xz) ^−1 +(xyz+y+xy) ^−1 +(1+z+yz)^−1

=(1+x+xz) ^−1 +y −1 (1+x+xz) ^−1 +(1+z+yz) ^−1

=(1+x+xz) ^−1 (1+y^ −1 )+(1+z+yz) ^−1

=(xyz+x+xz) ^−1 (1+y^−1 )+(1+z+yz) ^−1

=x ^−1 (1+z+yz)^-1 (1+y ^−1 )+(1+z+yz) ^−1

=(1+z+yz) ^−1 (x−1+(xy) ^−1 )+(1+z+yz) ^−1

=(1+z+yz) ^−1 (yz+z+1)

= 1+z+yz/1+z+yz

=1

= R.H.S.