Advertisements
Advertisements
Question
Prove that:
`("m"+"n")^-1("m"^-1+"n"^-1)=("m""n")^-1`
Theorem
Advertisements
Solution
`"L"."H"."S". ("m"+"n")^-1("m"^-1+"n"^-1)`
= `1/("m"+"n")(1/"m"+1/"n")`
= `1/("m"+"n").("n"+"m")/("m""n")`
= `1/("m""n"`
= `("m""n")^-1`
= R.H.S.
Hence, proved.
shaalaa.com
More About Exponents
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Compute:
`(56/28)^0÷(2/5)^3xx16/25`
Compute:
`(125)^(-2/3)÷(8)^(2/3)`
Simplify:
`8^(4/3)+25^(3/2)-(1/27)^(-2/3)`
Evaluate:
`[(10^3)^0]^5`
Evaluate:
`(7"x"^0)^2`
Simplify:
`("a"^5"b"^2)/("a"^2"b"^-3`
Simplify:
`(27"x"^-3"y"^6)^(2/3)`
Simplify:
`(-2"x"^(2/3)"y"^(-3/2))^6`
Simplify and express as positive indice:
(xy)(m-n).(yz)(n-l).(zx)(l-m)
Find the value of n, when:
`("a"^(2"n"-3)xx("a"^2)^("n"+1))/(("a"^4)^-3)=("a"^3)^3÷("a"^6)^-3`
