Question

rationalise the denominator of 1/√7-√6​

Answer 1

Answer:

$=>\sqrt{7} +\sqrt{6}$

Step-by-step explanation:

∵R.F. ( Rationalizing factor) of denominator $\sqrt{7} -\sqrt{6} =\sqrt{7}+ \sqrt{6}$

∴$\frac{1}{\sqrt{7}- \sqrt{6} } =\frac{1}{\sqrt{7} -\sqrt{6} } *\frac{\sqrt{7}+ \sqrt{6} }{\{\sqrt{7}+ \sqrt{6}}$

               $=\frac{\sqrt{7}+ \sqrt{6} }{7-6}$

∵$[(\sqrt{7} -\sqrt{6}) ( \sqrt{7} +\sqrt{6} )=(\sqrt{7}) ^{2} -(\sqrt{6}) ^{2} =7-6=1]$

              $=>\sqrt{7}+ \sqrt{6}$

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