Question

$\sqrt{11+6\sqrt{2} } +\sqrt{7+4\sqrt{3} } +\sqrt{5-2\sqrt{6} } =$

Answer 1

Răspuns:

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Explicație pas cu pas:

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Answer 2

$\sqrt{11+6\sqrt{2} } +\sqrt{7+4\sqrt{3} } +\sqrt{5-2\sqrt{6} } =\\\sqrt{(3+\sqrt{2} )^2} +\sqrt({2+\sqrt{3})^2 } +\sqrt{(\sqrt{2}-\sqrt{3})^2 } =\\3+\sqrt{2} +2+\sqrt{3} -\sqrt{2} +\sqrt{3} =\\3+2+\sqrt{3} +\sqrt{3} =\\5+2\sqrt{3}$

$a^2+2ab+b^2=(a+b)^2$

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