Advertisements
Advertisements
प्रश्न
Simplify \[\left[ \left\{ \left( 625 \right)^{- 1/2} \right\}^{- 1/4} \right]^2\]
Advertisements
उत्तर
We have to simplify. `[{(625)^((-1)/2)}^((-1)/4)]^2`So,
`[{(625)^((-1)/2)}^((-1)/4)]^2 = [{1/(625^(1/2))}^((-1)/4)\]^2`
`= [{1/5^(4 xx -1/2)}^((-1)/4)]^2`
`= [{1/5^2}^((-1)/4)]^2`
`= [{1/5^(2 xx 1/4)}]^2`
`[{(625)^((-1)/2)}^((-1)/4)]^2` `= [{1/5^((-1)/2)}]^2`
`[{1/(1/5^(1/2))}]`
`= [{1 xx 5^(1/2)}]^2`
`= 5^(1/2xx2)`
= 5
Hence, the value of `[{(625)^((-1)/2)}^((-1)/4)]^2` is 5.
APPEARS IN
संबंधित प्रश्न
Find:-
`64^(1/2)`
Simplify the following:
`(2x^-2y^3)^3`
Prove that:
`1/(1 + x^(b - a) + x^(c - a)) + 1/(1 + x^(a - b) + x^(c - b)) + 1/(1 + x^(b - c) + x^(a - c)) = 1`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
