if xyz=1 then show that (1+x+y^-1)^-1 + (1 + y+z^-1) + (1+z+x^-1) = 1
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Answered by
28
here is the answer....................
Attachments:
pandeytaapas:
in step 3 there is some error please check once again
and we cannot let the exact value of any given variable
we can
ultimately we use these equations with no.
otherwise what is the point
Answered by
18
Step-by-step explanation:
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.
:)
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