Advertisements
Advertisements
प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt2/sqrt3)^5(6/7)^2`
उत्तर
We have to simplify the following, assuming that x, y, z are positive real numbers
Given `(sqrt2/sqrt3)^5(6/7)^2`
`=(sqrt2/sqrt3)^(2+2+1)(6/7)^2`
`=(sqrt2/sqrt3)^2xx(sqrt2/sqrt3)^2xx(sqrt2/sqrt3)^1xx(6/7)^2`
`=(2/3)xx(2/3)xx(sqrt2/sqrt3)^1xx(6/7)^2`
`=(16sqrt2)/(49sqrt3)`
`=sqrt(512/7203)`
`=(512/7203)^(1/2)`
APPEARS IN
संबंधित प्रश्न
Prove that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
Given `4725=3^a5^b7^c,` find
(i) the integral values of a, b and c
(ii) the value of `2^-a3^b7^c`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
The square root of 64 divided by the cube root of 64 is
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
If a, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
If (16)2x+3 =(64)x+3, then 42x-2 =
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.