Question

Ajuda no desenvolvimento da expressão:

$( \sqrt{6} + \sqrt{2} )* \sqrt{2- \sqrt{3}} =$

Obrigado desde já ;D

Answer 1

$(\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}=\sqrt{2-\sqrt{3}}\cdot(\sqrt{6}+\sqrt{2})\\\\(\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}=\sqrt{2-\sqrt{3}}\cdot\sqrt{6}+\sqrt{2-\sqrt{3}}\cdot\sqrt{2}$

O produto das raízes é a raiz do produto:

$(\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}=\sqrt{6(2-\sqrt{3})}+\sqrt{2(2-\sqrt{3})}\\\\(\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}=\sqrt{12-6\sqrt{3}}+\sqrt{4-2\sqrt{3}}$

Vamos elevar os dois lados da igualdade ao quadrado:

$\left((\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}\right)^{2}=(\sqrt{12-6\sqrt{3}}+\sqrt{4-2\sqrt{3}})^{2}\\\\=(\sqrt{12-6\sqrt{3}})^{2}+2\sqrt{(12-6\sqrt{3})(4-2\sqrt{3})}+(\sqrt{4-2\sqrt{3}})^{2}\\\\=12-6\sqrt{3}+2\sqrt{48-24\sqrt{3}-24\sqrt{3}+12\cdot3}+4-2\sqrt{3}\\\\=16-8\sqrt{3}+2\sqrt{48+36-48\sqrt{3}}\\\\=16-8\sqrt{3}+2\sqrt{84-48\sqrt{3}}\\\\=16-8\sqrt{3}+2\sqrt{12(7-4\sqrt{3})}\\\\=16-8\sqrt{3}+2\sqrt{12}\sqrt{7-2\cdot\sqrt{3}\cdot2}$

O que vou fazer agora é completar o quadrado da última raiz:

$=16-8\sqrt{3}+2\sqrt{4\cdot3}\sqrt{7-2\cdot\sqrt{3}\cdot2+2^{2}-2^{2}}\\\\=16-8\sqrt{3}+2\sqrt{4}\sqrt{3}\sqrt{7-4-2\cdot\sqrt{3}\cdot2+2^{2}}\\\\=16-8\sqrt{3}+2\cdot2\sqrt{3}\sqrt{3-2\cdot\sqrt{3}\cdot2+2^{2}}\\\\=16-8\sqrt{3}+4\sqrt{3}\sqrt{(\sqrt{3})^{2}-2\cdot\sqrt{2}\cdot2+2^{2}}$

Podemos escrever o último radicando como um quadrado da diferença:

$=16-8\sqrt{3}+4\sqrt{3}\sqrt{(\sqrt{3}-2)^{2}}\\\\=16-8\sqrt{3}+4\sqrt{3}\sqrt{([-1]\cdot[2-\sqrt{3}])^{2}}\\\\=16-8\sqrt{3}+4\sqrt{3}\sqrt{(-1)^{2}(2-\sqrt{3})^{2}}\\\\=16-8\sqrt{3}+4\sqrt{3}\sqrt{(2-\sqrt{3})^{2}}\\\\=16-8\sqrt{3}+4\sqrt{3}(2-\sqrt{3})\\\\=16-8\sqrt{3}+8\sqrt{3}-4\cdot3\\\\=16-12\\\\=4$

Então, descobrimos que

$\left((\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}\right)^{2}=4$

Portanto:

$(\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}=\sqrt{4}\\\\\boxed{\boxed{(\sqrt{6}+\sqrt{2})\cdot\sqrt{2-\sqrt{3}}=2}}$

Answer 2

Para resolver esta questão, utilizaremos a fórmula para simplificar raízes aninhadas ("raiz dentro de raiz"):

$\boxed{ \begin{array}{c} \sqrt{a \pm b}=\sqrt{\dfrac{a+\sqrt{a^{2}-b^{2}}}{2}} \pm \sqrt{\dfrac{a-\sqrt{a^{2}-b^{2}}}{2}} \end{array} }$

onde $0 \leq b \leq a$.


Vamos tentar simplificar a expressão

$\sqrt{2-\sqrt{3}}$

utilizando a fórmula acima. Fazendo

$a=2\;\text{ e }\;b=\sqrt{3}$


$\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{2+\sqrt{2^{2}-\left(\sqrt{3} \right )^{2}}}{2}} - \sqrt{\dfrac{2-\sqrt{2^{2}-\left(\sqrt{3} \right )^{2}}}{2}}\\ \\ \sqrt{2-\sqrt{3}}=\sqrt{\dfrac{2+\sqrt{4-3}}{2}} - \sqrt{\dfrac{2-\sqrt{4-3}}{2}}\\ \\ \sqrt{2-\sqrt{3}}=\sqrt{\dfrac{2+\sqrt{1}}{2}} - \sqrt{\dfrac{2-\sqrt{1}}{2}}\\ \\ \sqrt{2-\sqrt{3}}=\sqrt{\dfrac{2+1}{2}} - \sqrt{\dfrac{2-1}{2}}\\ \\ \sqrt{2-\sqrt{3}}=\sqrt{\dfrac{3}{2}} - \sqrt{\dfrac{1}{2}}\\ \\ \sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3}-\sqrt{1}}{\sqrt{2}}\\ \\ \sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\\ \\ \sqrt{2-\sqrt{3}}=\dfrac{\left(\sqrt{3}-1 \right )\cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}\\ \\ \sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3}\cdot \sqrt{2}-1\cdot \sqrt{2}}{2}\\ \\ \sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3 \cdot 2}-\sqrt{2}}{2}$

$\boxed{ \begin{array}{c} \sqrt{2-\sqrt{3}}=\dfrac{\sqrt{6}-\sqrt{2}}{2} \end{array} }$


Então, desenvolvendo a expressão inicial, temos

$\left(\sqrt{6}+\sqrt{2} \right ) \cdot \sqrt{2-\sqrt{3}}\\ \\ =\left(\sqrt{6}+\sqrt{2} \right ) \cdot \left(\dfrac{\sqrt{6}-\sqrt{2}}{2} \right )\\ \\ =\dfrac{\left(\sqrt{6}+\sqrt{2} \right )\cdot \left(\sqrt{6}-\sqrt{2} \right )}{2}\\ \\ =\dfrac{\left(\sqrt{6} \right )^{2}-\left(\sqrt{2} \right )^{2}}{2}\\ \\ =\dfrac{6-2}{2}\\ \\ =\dfrac{4}{2}\\ \\ =2\\ \\ \\ \boxed{\left(\sqrt{6}+\sqrt{2} \right ) \cdot \sqrt{2-\sqrt{3}}=2}$