Advertisements
Advertisements
प्रश्न
The simplest rationalising factor of \[\sqrt[3]{500}\] is
पर्याय
\[\sqrt[3]{2}\]
\[\sqrt[3]{5}\]
\[\sqrt{3}\]
none of these
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
If 24 × 42 =16x, then find the value of x.
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x2 + y +y2 =
The positive square root of \[7 + \sqrt{48}\] is
