Question

rationalize the denominator ​

Answer 1

Answer:

Your answer attached in the photo

Answer 2

Step-by-step explanation:

✯1st Question:

Given:-

$\frac{ \sqrt{3} + 1}{2 \sqrt{2} - \sqrt{3} }$

What to do:-

To rationalise the denominator.

Solution:-

Let's solve the problem,

we have

$\frac{ \sqrt{3} + 1}{2 \sqrt{2} - \sqrt{3} }$

✯The denominator is 2√2-√3. Multiplying the numerator and denominator by 2√2+√3. We get,

$⟹ \frac{ \sqrt{3} + 1 }{2 \sqrt{2} - \sqrt{3} } \times \frac{2 \sqrt{2} + \sqrt{3} }{2 \sqrt{2} + \sqrt{3} }$

$⟹ \frac{( \sqrt{3} + 1)(2 \sqrt{2} + \sqrt{3}) }{(2 \sqrt{2} - \sqrt{3} )(2 \sqrt{2} + \sqrt{3} ) }$

⬤ Applying Algebraic Identity

(a-b)(a+b) = a² - b² to the denominator

We get,

$⟹ \frac{({ \sqrt{3} + 1)(2 \sqrt{2} + \sqrt{3}) } }{(2 \sqrt{2} {)}^{2} - ( \sqrt{3} {)}^{2} }$

$⟹ \frac{( \sqrt{3} + 1)(2 \sqrt{2} + \sqrt{3} )}{8 - 3}$

$⟹ \frac{( \sqrt{3} + 1)(2 \sqrt{2} + \sqrt{3} )}{5}$

Now,

Multiplying the expressions in the numerator left side to right side.

$⟹ \frac{2 \sqrt{6} + 3 + 2 \sqrt{2} + \sqrt{3} }{5}$

Hence, the denominator is rationalised.

Answer:-

$\frac{2 \sqrt{6} + 3 + 2 \sqrt{2} + \sqrt{3} }{5}$

Used Formulae:-

(a-b)(a+b) = a² - b²

✯2nd Question:

Given:

Now,

(3√2 + 1)/(2√5 - 3)

We rationalize the denominator by multiplying both the numerator and the denominator by conjugate rational number (2√5 + 3)

Then,

(3√2 + 1)/(2√5 - 3)

= {(3√2 + 1)(2√5 + 3)}/{(2√5 - 3)(2√5 + 3)}

= (6√10 + 9√2 + 2√5 + 3)/(20 - 9)

= (6√10 + 9√2 + 2√5 + 3)/11,

which is the required value.

In attachment I have answered this problem.⤴️⬆️

:)