Question

qual o resultado de $\sqrt{2} ( \sqrt{2+ \sqrt{2} } ) ( \sqrt{2- \sqrt{2} } )$

Answer 1

Usaremos as seguintes propriedades (de radiciação e produtos notáveis):

$\bullet\,\,\,\sqrt{a}\cdot\sqrt{b}=\sqrt{a\cdot b}\\\\\bullet\,\,\,(a+b)\cdot(a-b)=a^{2}-b^{2}$
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$\sqrt{2}\big(\sqrt{2+\sqrt{2}}\big)\big(\sqrt{2-\sqrt{2}}\big)$

Usando a propriedade 1 nas duas últimas raízes:

$\sqrt{2}\big(\sqrt{2+\sqrt{2}}\big)\big(\sqrt{2-\sqrt{2}}\big)=\sqrt{2}\cdot\sqrt{(2+\sqrt{2})(2-\sqrt{2})}$

Agora, tendo $a=2$ e $b=\sqrt{2}$, temos, dentro da segunda raiz, o seguinte produto notável:

$(2+\sqrt{2})(2-\sqrt{2})=(a+b)(a-b)=a^{2}-b^{2}$

Então:

$\sqrt{2}\big(\sqrt{2+\sqrt{2}}\big)\big(\sqrt{2-\sqrt{2}}\big)=\sqrt{2}\sqrt{2^{2}-(\sqrt{2})^{2}}\\\\\\\sqrt{2}\big(\sqrt{2+\sqrt{2}}\big)\big(\sqrt{2-\sqrt{2}}\big)=\sqrt{2}\sqrt{4-2}\\\\\\\sqrt{2}\big(\sqrt{2+\sqrt{2}}\big)\big(\sqrt{2-\sqrt{2}}\big)=\sqrt{2}\cdot\sqrt{2}\\\\\\\boxed{\boxed{\sqrt{2}\bigg(\sqrt{2+\sqrt{2}}\bigg)\bigg(\sqrt{2-\sqrt{2}}\bigg)=2}}$