Advertisements
Advertisements
Question
The simplest rationalising factor of \[\sqrt[3]{500}\] is
Options
\[\sqrt[3]{2}\]
\[\sqrt[3]{5}\]
\[\sqrt{3}\]
none of these
Advertisements
Solution
Given that: `3sqrt500` To find simplest rationalizing factor of the given expression we will factorize it as
`3sqrt500 = 3sqrt(125xx 4)`
`= 3sqrt(5xx5xx5xx 4)`
`= 3sqrt((5))^3 xx 3sqrt4`
` = 5 3sqrt4`
The rationalizing factor of `5 3sqrt4`is, `3sqrt2`since when we multiply given expression with this factor we get rid of irrational term.
Therefore, rationalizing factor of the given expression is `3sqrt2`
APPEARS IN
RELATED QUESTIONS
Simplify the following
`((4xx10^7)(6xx10^-5))/(8xx10^4)`
If `1176=2^a3^b7^c,` find a, b and c.
Simplify:
`(16^(-1/5))^(5/2)`
Simplify:
`root(lm)(x^l/x^m)xxroot(mn)(x^m/x^n)xxroot(nl)(x^n/x^l)`
Write \[\left( \frac{1}{9} \right)^{- 1/2} \times (64 )^{- 1/3}\] as a rational number.
If a, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
Find:-
`125^((-1)/3)`
Simplify:
`11^(1/2)/11^(1/4)`
