Question

The denominator after rationalisation  13√ +2√ of 13 +2 ​

Answer 1

Answer:

Answer:

\frac{4}{2+\sqrt{3}+\sqrt{7}}2+3+74 = \frac{(2\sqrt{3}+3-\sqrt{21})}{3}3(23+3−21)

Explanation:

Given \frac{4}{2+\sqrt{3}+\sqrt{7}}2+3+74

multiply numerator and denominator by (2+√3-√7), we get

=\frac{4(2+\sqrt{3}-\sqrt7)}{[(2+\sqrt{3})+\sqrt{7}][(2+\sqrt{3})-\sqrt{3}]}[(2+3)+7][(2+3)−3]4(2+3−7)

= \frac{4(2+\sqrt{3}-\sqrt{7})}{\left(2+\sqrt{3}\right)^{2}-\left(\sqrt{7}\right)^{2}}(2+3)2−(7)24(2+3−7)

= \frac{4(2+\sqrt{3}-\sqrt{7})}{2^{2}+2\times2\times\sqrt{3}+\left(\sqrt{3}\right)^{2}-7}22+2×2×3+(3)2−74(2+3−7)

=\frac{4(2+\sqrt{3}-\sqrt{7})}{4+4\sqrt{3}+3-7}4+43+3−74(2+3−7)

=\frac{4(2+\sqrt{3}-\sqrt{7})}{4\sqrt{3}}434(2+3−7)

=\frac{(2+\sqrt{3}-\sqrt{7})}{\sqrt{3}}3(2+3−7)

Rationalising the denominator, we get

= \frac{(2+\sqrt{3}-\sqrt{7})\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}3×3(2+3−7)×3

= \frac{(2\sqrt{3}+3-\sqrt{21})}{3}3(23+3−21)

Therefore,

\frac{4}{2+\sqrt{3}+\sqrt{7}}2+3+74 = \frac{(2\sqrt{3}+3-\sqrt{21})}{3}3(2

Step-by-step explanation:

l hope it will help you