Question

{1 / [√(9) - √(8)]} - {1 ÷ [√(8) - √(7)]} + {1 ÷ [√(7) - √(6]} - {1 ÷ [√(6) - √(5]} + {1 ÷ [√(5) - √(4]}

Answer 1

Please see the attachment

Answer 2

$\frac{1}{(\sqrt{9}-\sqrt{8})}-\frac{1}{(\sqrt{8}-\sqrt{7})}+\frac{1}{(\sqrt{7}-\sqrt{6})}-\frac{1}{(\sqrt{6}-\sqrt{5})}+\frac{1}{(\sqrt{5}-\sqrt{4})}=5$

Step-by-step explanation:

1. Rationalise the denominator of each term

$\frac{1}{(\sqrt{9}-\sqrt{8})}=\frac{1}{(\sqrt{9}-\sqrt{8})}\times \frac{(\sqrt{9}+\sqrt{8})}{(\sqrt{9}+\sqrt{8})}=\frac{(\sqrt{9}+\sqrt{8})}{((\sqrt{9})^{2}-(\sqrt{8})^{2})}=\frac{(\sqrt{9}+\sqrt{8})}{(9-8)}=\frac{(\sqrt{9}+\sqrt{8})}{(1)}=(\sqrt{9}+\sqrt{8})$     ...1)

$\frac{1}{(\sqrt{8}-\sqrt{7})}=\frac{1}{(\sqrt{8}-\sqrt{7})}\times \frac{(\sqrt{8}+\sqrt{7})}{(\sqrt{8}+\sqrt{7})}=\frac{(\sqrt{8}+\sqrt{7})}{((\sqrt{8})^{2}-(\sqrt{7})^{2})}=\frac{(\sqrt{8}+\sqrt{7})}{(8-7)}=\frac{(\sqrt{8}+\sqrt{7})}{(1)}=(\sqrt{8}+\sqrt{7})$     ...2)

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

$\frac{1}{(\sqrt{6}-\sqrt{5})}=\frac{1}{(\sqrt{6}-\sqrt{5})}\times \frac{(\sqrt{6}+\sqrt{5})}{(\sqrt{6}+\sqrt{5})}=\frac{(\sqrt{6}+\sqrt{5})}{((\sqrt{6})^{2}-(\sqrt{5})^{2})}=\frac{(\sqrt{6}+\sqrt{5})}{(6-5)}=\frac{(\sqrt{6}+\sqrt{5})}{(1)}=(\sqrt{6}+\sqrt{5})$     ...4)

$\frac{1}{(\sqrt{5}-\sqrt{4})}=\frac{1}{(\sqrt{5}-\sqrt{4})}\times \frac{(\sqrt{5}+\sqrt{4})}{(\sqrt{5}+\sqrt{4})}=\frac{(\sqrt{5}+\sqrt{4})}{((\sqrt{5})^{2}-(\sqrt{4})^{2})}=\frac{(\sqrt{5}+\sqrt{4})}{(5-4)}=\frac{(\sqrt{5}+\sqrt{4})}{(1)}=(\sqrt{5}+\sqrt{4})$     ...5)

2. from question

$\frac{1}{(\sqrt{9}-\sqrt{8})}-\frac{1}{(\sqrt{8}-\sqrt{7})}+\frac{1}{(\sqrt{7}-\sqrt{6})}-\frac{1}{(\sqrt{6}-\sqrt{5})}+\frac{1}{(\sqrt{5}-\sqrt{4})}$

$(\sqrt{9}+\sqrt{8})-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+\sqrt{4})$

$\sqrt{9}+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+\sqrt{4}$

$\sqrt{9}+\sqrt{4}=3+2=5$