Question

the rationalising factor of n√a/b

Answer 1

n√a/b the rationalizing factor is b
I. e n√a/b*b/b = nb√a/b*b
hope it helps you if so please mark it as brainliest

Answer 2

The rationalizing factor of $\sqrt[n]{\dfrac{a}{b}}$ is $\sqrt[n]{(\dfrac{a}{b})^{n-1}}$

Tips:

To find the rationalising factor of a fraction, we have to take such a term, by which when multiplied the given fraction, we obtain a rational number.

Step-by-step explanation:

Here, the given fraction is $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$

Clearly, its numerator is $\sqrt[n]{a}$ and the denominator is $\sqrt[n]{b}$, both of them are irrational numbers.

Let, the rationalising factor be $R$ Then

$\quad \sqrt[n]{\dfrac{a}{b}}\times R=\dfrac{a}{b}$

$\Rightarrow \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}\times R=\dfrac{a}{b}$

$\Rightarrow \dfrac{a^{\frac{1}{n}}}{b^{\frac{1}{n}}}\times R=\dfrac{a}{b}$

$\Rightarrow R=\dfrac{a^{1-\frac{1}{n}}} {b^{1-\frac{1}{n}}}$

$\Rightarrow R=\dfrac{a^{\frac{n-1}{n}}}{b^{\frac{n-1}{n}}}$

$\Rightarrow R=(\dfrac{a}{b})^{\frac{n-1}{n}}$

$\Rightarrow R=\sqrt[n]{(\dfrac{a}{b})^{n-1}}$

∴ the rationalising factor is $\sqrt[n]{(\dfrac{a}{b})^{n-1}}$