Advertisements
Advertisements
प्रश्न
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
विकल्प
\[\left\{ \left( \frac{5}{6} \right)^\frac{1}{5} \right\}^{- \frac{3}{6}}\]
\[\frac{1}{\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{1/6}}\]
\[\left( \frac{6}{5} \right)^{1/30}\]
\[\left( \frac{5}{6} \right)^{- 1/30}\]
Advertisements
उत्तर
We have to find the value of `{(5/6)^(1/5)}^((-1) / 6)`
So,
`{(5/6)^(1/5)}^((-1) / 6) = 5^(1/5 xx (-1)/6) / 6^(1/5 xx (-1)/6)`
`=5^(-1/30)/(6^((-1)/30))`
`=(1/(5^(-1/30))) / (1/(6^(1/30))`
`{(5/6)^(1/5)}^((-1) / 6)` = `1/(5^(1/30)) xx (6^(1/30))/1`
= `(6^(1/30))/5^(1/30)`
= `(6/5)^(1/30)`
APPEARS IN
संबंधित प्रश्न
Simplify:-
`2^(2/3). 2^(1/5)`
Solve the following equation for x:
`7^(2x+3)=1`
Solve the following equation for x:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
Find the value of x in the following:
`(sqrt(3/5))^(x+1)=125/27`
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, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
