Question

rationalize the denominator 1 divided by root 5 minus root 2 minus 1​

Answer 1

Answer:

On rationalizing the denominator of the denominator we get =\frac{\sqrt{6}+3\sqrt{5}+4\sqrt{15}-9\sqrt{2}}{39}

Step-by-step explanation:

\frac{1}{\sqrt{5}+\sqrt{6}-\sqrt{2}}

=\frac{1}{(\sqrt{5}+\sqrt{6})-\sqrt{2}}\times\frac{(\sqrt{5}+\sqrt{6})+\sqrt{2}}{(\sqrt{5}+\sqrt{6})+\sqrt{2}}

=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{(\sqrt{5}+\sqrt{6})^2-(\sqrt{2})^2}

=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{5+6+2\sqrt{30}-2}

=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{9+2\sqrt{30}}

=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{9+2\sqrt{30}}\times\frac{9-2\sqrt{30}}{9-2\sqrt{30}}

=\frac{(\sqrt{5}+\sqrt{6}+\sqrt{2})(9-2\sqrt{30})}{9^2-(2\sqrt{30})^2}

=\frac{9\sqrt{5}+9\sqrt{6}+9\sqrt{2}-10\sqrt{6}-12\sqrt{5}-4\sqrt{5}}{81-4\times30}

=\frac{9\sqrt{2}-\sqrt{6}-3\sqrt{5}-4\sqrt{15}}{-39}

=\frac{\sqrt{6}+3\sqrt{5}+4\sqrt{15}-9\sqrt{2}}{39}

Answer 2

Answer:

0

Step-by-step explanation: