Question

rationalize 3/√3+√7+2​

Answer 1

Answer:

$\frac{20+8\sqrt{3}-4\sqrt{21} }{48}$

Step-by-step explanation:

$\frac{3}{\sqrt{3}+\sqrt{7}+2 } \\\\=>\frac{3}{\sqrt{3}+(\sqrt{7}+2) }\\\\=>\frac{3}{\sqrt{3}+(\sqrt{7}+2) } \times \frac{\sqrt{3}-(\sqrt{7}+2) }{\sqrt{3}-(\sqrt{7}+2) } \\\\=>\frac{\sqrt{3}-(\sqrt{7}+2)}{{(\sqrt{3} )}^{2} - {(\sqrt{7}+2 )}^{2} } \\\\=>\frac{\sqrt{3}-\sqrt{7}-2}{3 - [{(\sqrt{7})}^{2}+2(\sqrt{7}) (2)+(2)^{2}] }\\\\=>\frac{\sqrt{3}-\sqrt{7}-2}{3 -( 7+4\sqrt{7}+4) }\\\\=>\frac{\sqrt{3}-\sqrt{7}-2}{3 - 7-4\sqrt{7}-4 }\\\\=>\frac{\sqrt{3}-\sqrt{7}-2}{-8-4\sqrt{7} }$

$\frac{(\sqrt{3}-\sqrt{7}-2)}{(-8-4\sqrt{7} )} \times \frac{(-8+4\sqrt{7})}{(-8+4\sqrt{7} )}\\\\=> \frac{-8(\sqrt{3}-\sqrt{7}-2){+4\sqrt{7}(\sqrt{3}-\sqrt{7}-2)}{}}{({-8)}^{2}-{(4\sqrt{7} )}^{2}}\\\\=> \frac{-8\sqrt{3}+8\sqrt{7}+16{+4\sqrt{21}-28-8\sqrt{7}}}{64-112}\\\\=>\frac{-20-8\sqrt{3}+4\sqrt{21} }{-48}\\\\ =>\frac{-(-20-8\sqrt{3}+4\sqrt{21} ) }{48}\\\\=>\frac{20+8\sqrt{3}-4\sqrt{21} }{48}$