Question

$A = 1/(\sqrt{2} +\sqrt{3} )\\B = 1/(\sqrt{3} -\sqrt{2} )\\\\\\Determine (A+B)^{2}$

Answer 1

Resposta:

$A = 1/(\sqrt{2} +\sqrt{3} )\\B = 1/(\sqrt{3} -\sqrt{2} )$

podemos racionalizar A e B para facilitar os cálculos, primeiro A:

$A = \frac{ \sqrt{2} - \sqrt{3} }{( \sqrt{2} + \sqrt{3}) ( \sqrt{2} - \sqrt{3}) } \\ A = \frac{ \sqrt{2} - \sqrt{3} }{2 - 3} \\ A = \sqrt{3} - \sqrt{2}$

agora B:

$B = \frac{1}{ \sqrt{3} -\sqrt{2} } = \\ B = \frac{ \sqrt{3} + \sqrt{2} }{3 - 2} = \sqrt{3} + \sqrt{2}$

agora, fazendo (A+B)^2:

$(A+B)^{2} = {( \sqrt{3} - \sqrt{2} + \sqrt{3} + \sqrt{2} )}^{2} \\ (A+B)^{2} = {(2 \sqrt{3} )}^{2} = 4 \times 3 = 12$