Question

√7-√8÷√5-√6 Rationalise the denominater
​

Answer 1

Answer:

$\frac{ \sqrt{7} - \sqrt{8} }{ \sqrt{5} - \sqrt{6} } \\ rationalising \: the \: denominater \: \\ \frac{ \sqrt{7} - \sqrt{8} }{ \sqrt{5} - \sqrt{6} } \times \frac{ \sqrt{5} + \sqrt{6} }{ \sqrt{5} + \sqrt{6} } \\ \frac{( \sqrt{7} - \sqrt{8} )( \sqrt{5} + \sqrt{6} )}{( \sqrt{5} ^{2} - \sqrt{6} ^{2}) } \\ = \frac{( \sqrt{7} - \sqrt{8} )( \sqrt{5} + \sqrt{6} ) }{ - 1}$

Answer 2

To Rationalize :- √7 - √8/√5 - √6

Used Concepts :- Rationalization.

A identity a² - b² = ( a + b ) ( a - b )

Important Points to remember in these questions :-

( i ) We can also rationalize numerator .

( ii ) We want to get correct answer . But we didn't know that we have to rationalize at numerator or denominator . So we have to rationalize at that whose either denominator or numerator is single.

Solution :-

$\frac{ \sqrt{7 } - \sqrt{8} }{ \sqrt{5} - \sqrt{6} }$

On Rationalising the denominator we get ,

$\frac{ \sqrt{7} - \sqrt{8} }{ \sqrt{5} - \sqrt{6} } \times \frac{ \sqrt{5} + \sqrt{6} }{ \sqrt{5} + \sqrt{6} }$

$\frac{ \sqrt{7}( \sqrt{5} + \sqrt{6}) - \sqrt{8}( \sqrt{5} + \sqrt{6}) }{5 - 6}$

$\frac{ \sqrt{35} + \sqrt{42} - \sqrt{40} - \sqrt{48} }{ - 1}$

After Taking Minus Common we get,

$\frac{ - ( \sqrt{40} + \sqrt{48} - \sqrt{42} - \sqrt{35} ) }{ - 1}$

After Cancelling minus we get ,

$\sqrt{40} + \sqrt{48} - \sqrt{42} - \sqrt{35}$

$2 \sqrt{10} + 4 \sqrt{3} - \sqrt{42} - \sqrt{35}$

Therefore, The required answer is 2√10 + 4√3 - √42 - √35.