Primeiro, temos:
[tex](a + b)^2 = a^2 + 2ab + b^2[/tex]
Então:
[tex](\sqrt{2} + \sqrt{3})^2 = (\sqrt{2})^2 + 2\sqrt{2} \sqrt{3} + (\sqrt{3} )^2[/tex]
[tex](\sqrt{2})^2 + 2\sqrt{2} \sqrt{3} + (\sqrt{3} )^2 = 2 + 2\sqrt{6} + 3 \\2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}[/tex]
Iremos utilizar a multiplicação pelo conjugado.
[tex]5 + 2\sqrt{6} + \frac{1}{5 + 2\sqrt{6} }\\5 + 2\sqrt{6} + (\frac{1}{5 + 2\sqrt{6} })(\frac{5-2\sqrt{6} }{5-2\sqrt{6} })\\5 + 2\sqrt{6} + (\frac{5-2\sqrt{6}}{(5 + 2\sqrt{6})(5-2\sqrt{6})})\\5 + 2\sqrt{6} + (\frac{5-2\sqrt{6}}{5^2 - (2\sqrt{6})^2})\\5 + 2\sqrt{6} + (\frac{5-2\sqrt{6}}{25 - (2\sqrt{6})(2\sqrt{6})})\\2\sqrt{6})(2\sqrt{6}) = 4*(\sqrt{6})^2 = 4 * 6 = 24\\5 + 2\sqrt{6} + (\frac{5-2\sqrt{6}}{25 - 24})\\5 + 2\sqrt{6} + (\frac{5-2\sqrt{6}}{1})\\5 + 2\sqrt{6} + 5-2\sqrt{6} = 10[/tex]
