Question

Rationalise:
(1)              4/(2+root3+root7)
(2)              4/(2root3+root5)

Answer 1

`(1) (4)/(2+√(3) +√(7)) \\ \\ or, (4)/(2+√(3) +√(7)) * (2 - √(3) -√(7))/(2-√(3)-√(7)) \\ \\ => \frac{ \sqrt[2]{3} - √(21)+3}{3} \\ \\ \\ (2) \frac{4}{\sqrt[2]{3} + √(5)} \\ \\ or, \frac{4}{\sqrt[2]{3} + √(5)} * \frac{\sqrt[2]{3}-√(5)}{\sqrt[2]{3}-√(5)} \\ \\ => \frac{\sqrt[8]{3}-\sqrt[4]{5}}{7}`

Answer 2

`(4)/(2+\sqrt3+\sqrt7)\cdot(2-(\sqrt3+\sqrt7))/(2-(\sqrt3+\sqrt7))=(8-4\sqrt3-4\sqrt7)/(2^2-(\sqrt3+\sqrt7)^2)=(8-4\sqrt3-4\sqrt7)/(4-3-2√(3\cdot7)-7)\\\\=(8-4\sqrt3-4\sqrt7)/(-6-2√(21))=(-2(2\sqrt3+2\sqrt7-4))/(-2(3+√(21)))=(2\sqrt3+2\sqrt7-4)/(3+√(21))\cdot(3-√(21))/(3-√(21))\\\\=(6\sqrt3-2√(63)+6\sqrt7-2√(147)-12+4√(21))/(3^2-(√(21))^2)=(6\sqrt3-2√(9\cdot7)+6\sqrt7-2√(49\cdot3)-12+4√(21))/(9-21)`


`=(6\sqrt3-6\sqrt7+6\sqrt7-14\sqrt3-12+4√(21))/(-12)=(-8\sqrt3+4√(21)-12)/(-12)=(-4(2\sqrt3-√(21)+3))/(-12)\\\\=(2\sqrt3-√(21)+3)/(3)`

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`(4)/(2\sqrt3+\sqrt5)\cdot(2\sqrt3-\sqrt5)/(2\sqrt3-\sqrt5)=(8\sqrt3-4\sqrt5)/((2\sqrt3)^2-(\sqrt5)^2)=(8\sqrt3-4\sqrt5)/(4\cdot3-5)=(8\sqrt3-4\sqrt5)/(12-5)\\\\=(8\sqrt3-4\sqrt5)/(7)`