Question

show that 1/(3 - root8) - 1/(root8 - root7) + 1/(root7 - root6) - 1/(root6 - root5) + 1/(root 5 - 2) = 5

Answer 1

$\mathbb{ SOLUTION }$ :

LHS

$= \frac{1}{(3 - \sqrt{8}) } - \frac{1}{( \sqrt{8} - \sqrt{7} )} + \frac{1}{( \sqrt{7} - \sqrt{6} )} - \frac{1}{( \sqrt{6} - \sqrt{5} )} + \frac{1}{ \sqrt{5} - 2 } \\ \\ \\ = \frac{1}{(3 - \sqrt{8} )} \times \frac{(3 + \sqrt{8}) }{(3 + \sqrt{8} )} - \frac{1}{( \sqrt{8} - \sqrt{7} ) } \times \frac{( \sqrt{8} + \sqrt{7} )}{( \sqrt{8} + \sqrt{7} )} + \frac{1}{( \sqrt{7} - \sqrt{6} )} \times \frac{( \sqrt{7} + \sqrt{6} )}{( \sqrt{7} + \sqrt{6} )} - \frac{1}{( \sqrt{6} - \sqrt{5} )} \times \frac{( \sqrt{6} + \sqrt{5} )}{( \sqrt{6} + \sqrt{5} )} + \frac{1}{( \sqrt{5} - 2 )} \times \frac{( \sqrt{5} + 2) }{( \sqrt{5} + 2) }$

[ by rationalisation ]

$= \frac{(3 + \sqrt{8}) }{ {(3)}^{2} - {(8)}^{2} } - \frac{( \sqrt{8} + \sqrt{7} )}{ {( \sqrt{8}) }^{2} - {( \sqrt{7}) }^{2} } + \frac{( \sqrt{7} + \sqrt{6} )}{ {( \sqrt{7} )}^{2} - {( \sqrt{6} )}^{2} } - \frac{( \sqrt{6} + \sqrt{5} )}{ {( \sqrt{6} )}^{2} - {( \sqrt{5} )}^{2} } + \frac{( \sqrt{5} + 2)}{ {( \sqrt{5}) }^{2} - {(2)}^{2} }$

[ $\therefore$ (a-b) (a+b) = a² - b² ]

$= \frac{3 + \sqrt{8} }{9 - 8} - \frac{ \sqrt{8} - \sqrt{7} }{8 - 7} + \frac{ \sqrt{7} + \sqrt{6} }{7 - 6} - \frac{ \sqrt{6} + \sqrt{5} }{6 - 5} + \frac{ \sqrt{5} - 2}{5 - 4}$


$= 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2 \\ \\ = 3 + 2 = 5 =RHS$

Hence,

LHS = RHS

(PROVED)

Answer 2

here is your answer OK Given..... ☺☺

1 / (3 - root 8) = 3 + root 8

1 / (root 8 - root 7) = root 8 + root 7

1 / (root 7 - root 6) = root 7 + root 6

1 / (root 6 - root 5 ) = root 6 + root 5

1 / (root 5 -2) = root 5 + 2

3 + root 8 - root 8 - root 7 + root 7 + root 6 - root 6 - root 5 + root 5 + 2

3 + 2 = 5

Hence Proved

OK