Question

Rationalize the denominator 1/ root 7− root 6 , 1/ root 5 +root 2 , 1 / root 7 − root 2 and 2/ root 3 −root 5 .

Answer 1

Dear student,

$$\begin{gathered} i)\frac{{\displaystyle 1}}{{\displaystyle \sqrt{7}-\sqrt{6}}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}{\displaystyle +}\sqrt{6}}{\sqrt{7}+\sqrt{6}} \\ =\frac{\sqrt{7}{\displaystyle +}\sqrt{6}}{{\displaystyle {\left(\sqrt{7}\right)}^2-{\left(\sqrt{6}\right)}^2}}=\frac{{\displaystyle \sqrt{7}+\sqrt{6}}}{7-6}=\sqrt{7}+\sqrt{6} \\ iv)\frac{2}{\sqrt{3}{\displaystyle -}{\displaystyle \sqrt{5}}}=\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}-\sqrt{5}}}\times \frac{\sqrt{3}{\displaystyle +}\sqrt{5}}{\sqrt{3}{\displaystyle +}\sqrt{5}} \\ =\frac{2(\sqrt{3}+\sqrt{5})}{\left(\sqrt{3}\right){\displaystyle 2}{\displaystyle -}{\displaystyle \left(\sqrt{5}\right)}{\displaystyle 2}}=\frac{2(\sqrt{3}+\sqrt{5})}{3-5}=\frac{2(\sqrt{3}+\sqrt{5})}{-2}=-(\sqrt{3}+\sqrt{5}) \end{gathered}$$

Regards