Rationalize the denominators of the following- 1-7-6 -4 MARKS-

Rationalisation : 2 Marks Calculation : 2 Marks 1/√(7) √(6)=1(√(7)+√(6))/(√(7) √(6))(√(7)+√(6)) =√(7)+√(6)/(√(7))^2 (√(6))^2=√(7)+√(6)/1=√(7)+√(6) ...

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Standard IX

Mathematics

Rationalisation

Rationalize t...

Question

Rationalize the denominators of the following:

$\frac{1}{(\sqrt{7} - \sqrt{6})}$ [4 MARKS]

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Solution

Rationalisation : 2 Marks
Calculation : 2 Marks

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

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Rationalize the denominators of the following: 1(√7−√6) [4 MARKS]

Rationalisation : 2 Marks Calculation : 2 Marks 1√7−√6=1(√7+√6)(√7−√6)(√7+√6)=√7+√6(√7)2−(√6)2=√7+√61=√7+√6

Rationalize the denominators of the following:

$\frac{1}{(\sqrt{7} - \sqrt{6})}$ [4 MARKS]

Rationalisation : 2 Marks
Calculation : 2 Marks

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