Question

rationalize the denominator of 1/√3-√2​

Answer 1

Answer:

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

Answer 2

☆ To Rationalize,

• Thebdenominator of $\sf\large\frac{1}{sqrt{3} - sqrt{2}}$

☆ Concept Used:-

$\sf \mapsto \frac{a}{a} = 1 \\ \\\sf \mapsto \: \frac{x}{a} \times \frac{a}{a} = \frac{x}{a}$

☆ ID Used:-

$\sf \leadsto \: (a + b)(a - b) = {a}^{2} - {b}^{2}.$

☆ Now,

$\sf\implies \frac{1}{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{1}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{1( \sqrt{3} + \sqrt{2}) }{( \sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2}) } \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{( \sqrt{3}) {}^{2} - ( \sqrt{2}) {}^{2} } \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{3 - 2} \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{1} \\ \\ \large \fbox\pink{ = \sqrt{3} + \sqrt{2} .}$