Advertisements
Advertisements
प्रश्न
Simplify the following:
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
Advertisements
उत्तर
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
`=(5xx(5^2)^(n+1)-(5^2)xx5^(2n))/(5xx5^(2n+3)-(5^2)^(n+1))`
`=(5xx(5^(2n+2))-(5^2)xx5^(2n))/(5xx5^(2n+3)-5^(2n+2))`
`=(5^(1+2n+2)-5^(2+2n))/(5^(1+2n+3)-5^(2n+2))`
`=(5^(2+2n)(5-1))/(5^(2+2n)(5^2-1))`
`=(5-1)/((5^2)-1)`
`=(5-1)/(25-1)`
`=4/24`
`=1/6`
APPEARS IN
संबंधित प्रश्न
Simplify the following
`(a^(3n-9))^6/(a^(2n-4))`
Solve the following equation for x:
`2^(5x+3)=8^(x+3)`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Simplify:
`(16^(-1/5))^(5/2)`
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
When simplified \[( x^{- 1} + y^{- 1} )^{- 1}\] is equal to
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
If 10x = 64, what is the value of \[{10}^\frac{x}{2} + 1 ?\]
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
Find:-
`32^(1/5)`
