Question

1 upon √6-√5 rationalise the denominator​

Answer 1

Answer:

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According to question:-

$\frac{1}{ \sqrt{6} - \sqrt{5} }$

Rationalising factor= √6+√5

$\frac{1}{ \sqrt{6} - \sqrt{5} } \times \frac{ \sqrt{6} + \sqrt{5} }{ \sqrt{6} + \sqrt{5} }$

$\frac{ \sqrt{6} + \sqrt{5} }{( { \sqrt{6} })^{2} - ( { \sqrt{5} })^{2} }$

$\frac{ \sqrt{6} + \sqrt{5} }{6 - 5}$

$\sqrt{6} + \sqrt{5}$

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Answer 2

Question :-

$\frac{1}{ \sqrt{6} - \sqrt{5} }$

What to do = rationalisation?

Solution :-

$\frac{1}{ \sqrt{6} - \sqrt{5} }$

Now multiply by in the numerator and the denominator

$( \sqrt{6} + \sqrt{5} )$

Hence,

$( \frac{1}{ \sqrt{6} - \sqrt{5} } ) \times ( \frac{ \sqrt{6} + \sqrt{5} }{ \sqrt{6} + \sqrt{5} } ) \\ \\ = ( \frac{ \sqrt{6} + \sqrt{5} }{6 - 5} ) = \sqrt{6} + \sqrt{5} \\ \\ = 2.44 + 2.23 =4.6 7 \: \: \: ans$

Formulae used :-

1)$( \sqrt{x} ) {}^{2} = x$

Example :-

$( \sqrt{7} ) {}^{2} = 7$

2)$(x - y)(x + y) = {x}^{2} - y {}^{2}$