Find the value of (4^-1 + 3^-1 + 6^-2)^-1 and (2^-1 + 3^-1 + 4^-1)^0
Can you tell me the answer in detail with steps.
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(4^-1 + 3 ^-1+6^-2)^-1
=(1/4+1/3+1/6^2)^-1
=(1/4+1/3+1/36)^-1
=[( 9 + 12 + 1 )/36]^-1
= (22/36)^-1
=(11/18)^-1
=18/11
(2^-1+3^-1+4^-1)^0
=1 (anything to the power0=1,x^0=1)
=(1/4+1/3+1/6^2)^-1
=(1/4+1/3+1/36)^-1
=[( 9 + 12 + 1 )/36]^-1
= (22/36)^-1
=(11/18)^-1
=18/11
(2^-1+3^-1+4^-1)^0
=1 (anything to the power0=1,x^0=1)
Gangavaram:
Ok thanks
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