Question

Rationalise the denominator of the following 1/root 7-root 6

Answer 1

Given :-

$\dfrac{1}{\sqrt{7}-\sqrt{6}}$

To find :-

Rationalise the denominator.

Solution:-

=$\dfrac{1}{ \sqrt{7} - \sqrt{6} } \times \dfrac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} + \sqrt{6} }$

=$\dfrac{1( \sqrt{7} + \sqrt{6} )}{ ( \sqrt{7 } - \sqrt{6} )\times( \sqrt{7} + \sqrt{6} )}$

using suitable identity,

$\boxed{\sf{ (a + b)(a - b) = {a}^{2} - {b}^{2} }}$

Now,

=$\dfrac{ \sqrt{7} + \sqrt{6} }{( {7)}^{2} - (6)^{2} }$

=$\dfrac{ \sqrt{7} + \sqrt{6} }{7- 6}$

=$\dfrac{ \sqrt{7} + \sqrt{6} }{1}$

hence, the required answer is

$= { \sqrt{7} + \sqrt{6} }{}$

Answer 2

Step-by-step explanation:

Solution:

1/(√7 - √6)

The denominator is √7 - √6.

We know that

Rationalising factor of √a - √b = √a + √b.

So, the rationalising factor of √7 - √6 = √7 + √6.

On comparing the denominator them

=> [1/(√7 - √6)]×[(√7 - √6)/(√7- √6)]

=> [1(√7 + √6)]/[(√7 - √6)(√7 + √6)]

Applying algebraic identity in denominator; (a-b)(a+b) = a^2 - b^2. Where, a = √7 and b = √6.

=> [1(√7 + √6)]/[(√7)^2 - (√6)^2)]

=> [1(√7 + √6)]/(7 - 6)

=> [1(√7 + √6)]/1

=> 1(√7 + √6)

=> √7 + √6

Hence, the denominator is rationalised. of

Answer:

→ √7 - √6

Used Formulae:

• Rationalising factor of √a - √b = √a + √b.

• (a-b)(a+b) = a^2 - b^2.