Advertisements
Advertisements
प्रश्न
Simplify the following:
`(5xx25^(n+1)-25xx5^(2n))/(5xx5^(2n+3)-25^(n+1))`
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)`
`=4/24`
`=1/6`
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Prove that:
`((0.6)^0-(0.1)^-1)/((3/8)^-1(3/2)^3+((-1)/3)^-1)=(-3)/2`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
Write the value of \[\sqrt[3]{125 \times 27}\].
For any positive real number x, write the value of \[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\]
The square root of 64 divided by the cube root of 64 is
Which one of the following is not equal to \[\left( \frac{100}{9} \right)^{- 3/2}\]?
If x = \[\frac{2}{3 + \sqrt{7}}\],then (x−3)2 =
Find:-
`32^(2/5)`
Simplify:
`11^(1/2)/11^(1/4)`
