Question

Rationalise the denominator of each of the following: √3+1/2√2-√3​

Answer 1

Answer:

3+1=4+2=6+2=8+3=11

Step-by-step explanation:

is the answer

Answer 2

Step-by-step explanation:

As per the provided information in the given question, we have to rationalise the denominator of :

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

In order to rationalise the denominator of any fraction, we multiply the rationalising factor of the denominator with both the numerator and denominator of fraction.

○ Rationalising factor is nothing but the conjugate of the denominator. Here, the denominator is in the form of (a – b). So, the rationalising factor will be (a + b). Henceforth, rationalising factor here will be (2√2 + √3).

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

Multiplying (2√2 + √3) with both the numerator and the denominator of the fraction.

$\longrightarrow \sf{\quad { \dfrac{(\sqrt{3} + 1)(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2} -\sqrt{3})(2\sqrt{2} + \sqrt{3})} }}$

Using the distributive property in the numerator to perform multiplication easily. And, in the denominator using the property,

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

$\longrightarrow \sf{\quad { \dfrac{\sqrt{3}(2\sqrt{2} + \sqrt{3}) + 1(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2})^2 -(\sqrt{3})^2} }}$

Simplifying further by performing multiplication in the numerator and writing the squares of the terms in the denominator.

$\longrightarrow \sf{\quad { \dfrac{2\sqrt{6} + 3 + 2\sqrt{2} + \sqrt{3} }{(4 \times 2) - 3} }}$

Performing multiplication in the denominator.

$\longrightarrow \sf{\quad { \dfrac{2\sqrt{6} + 3 + 2\sqrt{2} + \sqrt{3} }{8- 3} }}$

Performing subtraction in the denominator.

$\longrightarrow \quad \underline{\boxed { \dfrac{ \textbf{\textsf{2}}\sqrt{\textbf{\textsf{6}}} + \textbf{\textsf{3}} + \textbf{\textsf{2}}\sqrt{\textbf{\textsf{2}}} + \sqrt{\textbf{\textsf{3}}} }{\textbf{\textsf{5}}} }}$

$\underline { \sf { Hence, \; rationalised!! }}$