Question

Ajutati-ma la acest exercitiu, va rog

Answer 1

a)

$\frac{1}{2\sqrt[]{3}-3\sqrt{7}}=\frac{2\sqrt{3}+3\sqrt{7}  }{(2\sqrt{3}-3\sqrt{7})(2\sqrt{3} +3\sqrt{7} )}=\frac{2\sqrt{3}+3\sqrt{7}}{(2\sqrt{3} )^{2}-(3\sqrt{7} ){^2} }\\ =\frac{2\sqrt{3} +3\sqrt{7} }{12-63}=\frac{2\sqrt{3}+3\sqrt{7}  }{-51}$

b) $\frac{1}{2-\sqrt{3} } =\frac{2+\sqrt{3} }{(2-{\sqrt{3})(2+\sqrt{3} ) } } =\frac{2+\sqrt{3} }{4-3} =2+\sqrt{3}$