If a - 1-5 - -3- b - 1-7 - -5 and c - 1-9 - -7- then a - b - c equals

a = 1√(5) nbsp; √(3)×√(5) nbsp; + √(3)√(5) nbsp; + √(3) = √(5) nbsp; + √(3)5 3 = √(5) nbsp; + √(3)2 b = 1√(7) nbsp; + √(5)×√(7) nbsp ...

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Byju's Answer

Standard IX

Mathematics

Operations on Rational and Irrational Numbers

If a = 1/√5...

Question

If $a = \frac{1}{\sqrt{5} - \sqrt{3}}, b = \frac{1}{\sqrt{7} + \sqrt{5}}$ and $c = \frac{1}{\sqrt{9} - \sqrt{7}},$ then $a + b + c$ equals

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Solution

$a = \frac{1}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{\sqrt{5} + \sqrt{3}}{5 - 3} = \frac{\sqrt{5} + \sqrt{3}}{2}$
$b = \frac{1}{\sqrt{7} + \sqrt{5}} \times \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} - \sqrt{5}} = \frac{\sqrt{7} - \sqrt{5}}{7 - 5} = \frac{\sqrt{7} - \sqrt{5}}{2}$
$c = \frac{1}{\sqrt{9} + \sqrt{7}} \times \frac{\sqrt{9} - \sqrt{7}}{\sqrt{9} - \sqrt{7}} = \frac{\sqrt{9} - \sqrt{7}}{9 - 7} = \frac{\sqrt{9} - \sqrt{7}}{2}$
$∴ a + b + c$
$= \frac{\sqrt{5} + \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + \frac{\sqrt{9} - \sqrt{7}}{2}$
$= \frac{\sqrt{5} + \sqrt{3} + \sqrt{7} - \sqrt{5} + \sqrt{9} - \sqrt{7}}{2}$
$= \frac{\sqrt{3} + 3}{2}$
$∴ \frac{3 + \sqrt{3}}{2}$

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If a=1√5−√3,b=1√7+√5 and c=1√9−√7, then a+b+c equals

a=1√5−√3×√5+√3√5+√3=√5+√35−3=√5+√32b=1√7+√5×√7−√5√7−√5=√7−√57−5=√7−√52c=1√9+√7×√9−√7√9−√7=√9−√79−7=√9−√72∴a+b+c=√5+√32+√7−√52+√9−√72=√5+√3+√7−√5+√9−√72=√3+32∴3+√32

If $a = \frac{1}{\sqrt{5} - \sqrt{3}}, b = \frac{1}{\sqrt{7} + \sqrt{5}}$ and $c = \frac{1}{\sqrt{9} - \sqrt{7}},$ then $a + b + c$ equals