Advertisements
Advertisements
प्रश्न
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
विकल्प
5a2bc2
25ab2c
5a3bc3
125a2bc2
Advertisements
उत्तर
Find value of \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\]
\[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] = `5sqrt(5^5 a^10 b^5 c^10)`
`= 5^(5 xx 1/5) a^(10 xx 1/5 ) b^(5 xx 1/5 ) c^(10xx1/5)`
`= 5^(5 xx 1/5) a^(10 xx 1/5 ) b^(5 xx 1/5 ) c^(10xx1/5)`
\[\sqrt[5]{3125 a^{10} b^5 c^{10}} = 5 a^2 b c^2\]
APPEARS IN
संबंधित प्रश्न
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
Find the value of x in the following:
`2^(x-7)xx5^(x-4)=1250`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
The product of the square root of x with the cube root of x is
When simplified \[( x^{- 1} + y^{- 1} )^{- 1}\] is equal to
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
The simplest rationalising factor of \[\sqrt[3]{500}\] is
Simplify:
`11^(1/2)/11^(1/4)`
