Question

Simplificação de radicais / raízes quadradas.

Simplifique a expressão:

$\sqrt{223+21\sqrt{5}}+\sqrt{223-21\sqrt{5}}.$​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{x = \sqrt{223 + 21\sqrt{5}} + \sqrt{223 - 21\sqrt{5}}}$

$\mathsf{x^2 = (\sqrt{223 + 21\sqrt{5}} + \sqrt{223 - 21\sqrt{5}}\:)^2}$

$\mathsf{x^2 = (223 + 21\sqrt{5}) + 2\sqrt{49.729 - 2.205} + (223 - 21\sqrt{5})}$

$\mathsf{x^2 = 446 + 2\sqrt{47.524}}$

$\mathsf{x^2 = 446 + 2(218)}$

$\mathsf{x^2 = 446 + 436}$

$\mathsf{x^2 = 882}$

$\mathsf{x = \sqrt{2.3^2.7^2}}$

$\mathsf{x = 21\sqrt{2}}$

$\boxed{\boxed{\mathsf{\left(\sqrt{223 + 21\sqrt{5}} + \sqrt{223 - 21\sqrt{5}}\:\right) = 21\sqrt{2}}}}$

Answer 2

$\large\boxed{\begin{array}{l}\underline{\sf Radical\,duplo}\\\sf \sqrt{A\pm\sqrt{B}}=\sqrt{\dfrac{A+C}{2}}\pm\sqrt{\dfrac{A-C}{2}}\\\sf C=\sqrt{A^2-B}\\\sf 21\sqrt{5}=\sqrt{21^2\cdot5}=\sqrt{2205}\\\sf \sqrt{223+21\sqrt{5}}=\sqrt{223+\sqrt{2205}}\\\sf C=\sqrt{223^2-2205}=\sqrt{49729-2205}\\\sf C=\sqrt{47524}=218\\\sf \sqrt{223+\sqrt{2205}}=\sqrt{\dfrac{223+218}{2}}+\sqrt{\dfrac{223-218}{2}}\\\\\sf\sqrt{223+\sqrt{2205}}=\sqrt{\dfrac{441}{2}}+\sqrt{\dfrac{5}{2}}\end{array}}$

$\large\boxed{\begin{array}{l}\sf\sqrt{223+\sqrt{2205}}=\dfrac{21}{\sqrt{2}}+\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{21\sqrt{2}}{2}+\dfrac{\sqrt{10}}{2}\\\\\sf\sqrt{223+\sqrt{2205}}=\dfrac{21\sqrt{2}+\sqrt{10}}{2}\\\\\sf\sqrt{223-\sqrt{2205}}=\dfrac{21\sqrt{2}-\sqrt{10}}{2}\end{array}}$

$\small\boxed{\begin{array}{l}\sf\sqrt{223+21\sqrt{5}}+\sqrt{223-21\sqrt{5}}=\dfrac{21\sqrt{2}+\diagup\!\!\!\sqrt{10}+21\sqrt{2}-\diagup\!\!\!\sqrt{10}}{2}\\\\\sf\sqrt{223+21\sqrt{5}}+\sqrt{223-21\sqrt{5}}=\dfrac{\backslash\!\!\!2\cdot21\sqrt{2}}{\backslash\!\!\!2}\\\\\Large\boxed{\boxed{\boxed{\boxed{\sf \sqrt{223+21\sqrt{5}}+\sqrt{223-21\sqrt{5}}=21\sqrt{2}}}}}\end{array}}$