Question

rationalise √3+1÷2√2-√3​

Answer 1

Given:

The expression -

$\bf \dashrightarrow \: \dfrac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}}$

What To Find:

We have to

• Rationalise the denominator.

How To Find:

To find it, we have to

• First, find the conjugate or of the denominator.
• Next, multiply the conjugate with both the denominator and numerator.
• Then, simplify the expression using identities according.
• Finally, if the denominator is a rational no. then, it is rationalised.

Solution:

$\sf \longrightarrow \: \dfrac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}}$

Here the conjugate of the denominator is,

$\sf \longrightarrow \: 2\sqrt{2} + \sqrt{3}$

Multiply the conjugate with the expression,

$\sf \longrightarrow \: \dfrac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}} \times \sf \dfrac{2\sqrt{2} + \sqrt{3}}{2\sqrt{2} + \sqrt{3}}$

Also written as,

$\sf \longrightarrow \: \dfrac{\sqrt{3} + 1 \times 2\sqrt{2} + \sqrt{3}}{2\sqrt{2} - \sqrt{3} \times 2\sqrt{2} + \sqrt{3}}$

Multiply the numerator,

$\sf \longrightarrow \: \dfrac{4 \times 2\sqrt{2}}{2\sqrt{2} - \sqrt{3} \times 2\sqrt{2} + \sqrt{3}}$

Using the identity (a - b) (a + b) = a² + b²,

$\sf \longrightarrow \: \dfrac{4 \times 2\sqrt{2}}{(2\sqrt{2})^2 - (\sqrt{3})^2}$

Solve the denominator,

$\sf \longrightarrow \: \dfrac{4 \times 2\sqrt{2}}{8 - 3}$

Subtract 3 from 8,

$\sf \longrightarrow \: \dfrac{4 \times 2\sqrt{2}}{5}$

Final Answer:

∴ Thus, the answer is $\bf \dfrac{4 \times 2\sqrt{2}}{5}$ after rationalising the denominator.

Answer 2

Solution:

=(2√6+3+2√2+√3)/5

Step-by-Step Explaination:

=[(√3+1)/(2√2-√3)]×[(2√2+√3)/(2√2+√3)]

=[(√3+1)(2√2+√3)]/[(2√2)²-(√3)²]

=[√3(2√2+√3)+1(2√2+√3)]/[4(2)-3]

=(2√6+√3²+2√2+√3]/8-3

=(2√6+3+2√2+√3)/5