Question

The rationalising factor of 7√a⁴b³c⁵ is ?​

Answer 1

Step-by-step explanation:

$\sqrt[7]{ {a}^{4}. {b}^{3} \times {c}^{5} } \\ = {a}^{4 \times \frac{1}{7} } . {b}^{3 \times \frac{1}{7} } . {c}^{5 \times \frac{1}{7 } } \\ = {a}^{ \frac{4}{7} } . {b}^{ \frac{3}{7} } . {c}^{ \frac{5}{7} }$

hope it helps you

Answer 2

Answer:

Step-by-step explanation:

Given :    7 root a⁴b³c⁵

$\sqrt[7]{a^4b^3c^5}$

To Find : rationalizing factor

Solution :

$\sqrt[7]{a^4b^3c^5}$

Rationalizing factor :

The factor of multiplication by which an irrational number is multiplied to convert it into rational number

If the product of two irrational numbers or surds is a rational number, then each surd is a rationalizing factor for each other.

as there is 7th root so we need to have power of a , b and c as 7 inside 7th root

a⁷/a⁴ = a³

b⁷/b³ = b⁴

c⁷/c⁵ = c²

so rationalizing factor is

$\sqrt[7]{a^3b^4c^2}$

 

$\sqrt[7]{a^4b^3c^5} * \sqrt[7]{a^3b^4c^2} = abc$

option b) is correct

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