Verified answer

Resposta:

[tex]\textsf{Leia abaixo}[/tex]

Explicação passo a passo:

[tex]\sf \dfrac{3}{\sqrt{7 - 2\sqrt{10}}} + \dfrac{4}{\sqrt{8 + 4\sqrt{3}}} = \dfrac{1}{\sqrt{11 - 2\sqrt{30}}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{2^2\:.\:10}}} + \dfrac{4}{\sqrt{8 + \sqrt{4^2\:.\:3}}} = \dfrac{1}{\sqrt{11 - \sqrt{2^2\:.\:30}}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{4\:.\:10}}} + \dfrac{4}{\sqrt{8 + \sqrt{16\:.\:3}}} = \dfrac{1}{\sqrt{11 - \sqrt{4\:.\:30}}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} + \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \dfrac{1}{\sqrt{11 - \sqrt{120}}}[/tex]

[tex]\boxed{\sf \sqrt{A \pm \sqrt{B}} = \sqrt{\dfrac{A + C}{2}} \pm \sqrt{\dfrac{A - C}{2}} }[/tex]

[tex]\boxed{\sf C = A^2 - B}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}}\rightarrow\begin{cases}\sf A = 7\\\sf B = 40\\\sf C = \sqrt{49 - 40} = \sqrt{9} = 3\end{cases}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} = \dfrac{3}{\sqrt{\dfrac{7 + 3}{2}} - \sqrt{\dfrac{7 - 3}{2}}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} = \dfrac{3}{\sqrt{\dfrac{10}{2}} - \sqrt{\dfrac{4}{2}}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} = \dfrac{3}{\sqrt{5} - \sqrt{2}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} = \dfrac{3}{\sqrt{5} - \sqrt{2}}\:.\:\dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} = \dfrac{3\sqrt{5} + 3\sqrt{2}}{5 - 2}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} = \dfrac{3\sqrt{5} + 3\sqrt{2}}{3}[/tex]

[tex]\boxed{\sf \dfrac{3}{\sqrt{7 - \sqrt{40}}} = \sqrt{5} + \sqrt{2}}[/tex]

[tex]\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}}\rightarrow\begin{cases}\sf A = 8\\\sf B = 48\\\sf C = \sqrt{64 - 48} = \sqrt{16} = 4\end{cases}[/tex]

[tex]\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \dfrac{4}{\sqrt{\dfrac{8 + 4}{2}} + \sqrt{\dfrac{8 - 4}{2}}}[/tex]

[tex]\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \dfrac{4}{\sqrt{\dfrac{12}{2}} + \sqrt{\dfrac{4}{2}}}[/tex]

[tex]\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \dfrac{4}{\sqrt{6} + \sqrt{2}}[/tex]

[tex]\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \dfrac{4}{\sqrt{6} + \sqrt{2}}\:.\:\dfrac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}[/tex]

[tex]\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \dfrac{4\sqrt{6} - 4\sqrt{2}}{6 - 2}[/tex]

[tex]\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \dfrac{4\sqrt{6} - 4\sqrt{2}}{4}[/tex]

[tex]\boxed{\sf \dfrac{4}{\sqrt{8 + \sqrt{48}}} = \sqrt{6} - \sqrt{2}}[/tex]

[tex]\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}}\rightarrow\begin{cases}\sf A = 11\\\sf B = 120\\\sf C = \sqrt{121 - 120} = \sqrt{1} = 1\end{cases}[/tex]

[tex]\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}} = \dfrac{1}{\sqrt{\dfrac{11 + 1}{2}} - \sqrt{\dfrac{11 - 1}{2}}}[/tex]

[tex]\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}} = \dfrac{1}{\sqrt{\dfrac{12}{2}} - \sqrt{\dfrac{10}{2}}}[/tex]

[tex]\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}} = \dfrac{1}{\sqrt{6} - \sqrt{5}}[/tex]

[tex]\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}} = \dfrac{1}{\sqrt{6} - \sqrt{5}}\:.\:\dfrac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}}[/tex]

[tex]\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}} = \dfrac{\sqrt{6} + \sqrt{5}}{6 - 5}[/tex]

[tex]\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}} = \dfrac{\sqrt{6} + \sqrt{5}}{1}[/tex]

[tex]\boxed{\sf \dfrac{1}{\sqrt{11 - \sqrt{120}}} = \sqrt{6} + \sqrt{5}}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - 2\sqrt{10}}} + \dfrac{4}{\sqrt{8 + 4\sqrt{3}}} = \dfrac{1}{\sqrt{11 - 2\sqrt{30}}} \Leftrightarrow (\sqrt{5} + \sqrt{2}) + (\sqrt{6} - \sqrt{2}) = \sqrt{6} + \sqrt{5}[/tex]

[tex]\sf \dfrac{3}{\sqrt{7 - 2\sqrt{10}}} + \dfrac{4}{\sqrt{8 + 4\sqrt{3}}} = \dfrac{1}{\sqrt{11 - 2\sqrt{30}}} \Leftrightarrow \boxed{\boxed{\sf \sqrt{6} + \sqrt{5} = \sqrt{6} + \sqrt{5}}}[/tex]

Resposta:

[tex]\sf \sqrt{5}+\sqrt{6} =\sqrt{6}+\sqrt{5}[/tex]

Explicação passo a passo:

[tex]\sf \dfrac{3}{\sqrt{7-2\sqrt{10} } } + \dfrac{4}{\sqrt{8+4\sqrt{3} } } =\dfrac{1}{\sqrt{11-2\sqrt{30} } } \\ \\ \\ \\[/tex]

Resolvendo a primeira fração:

[tex]\sf \dfrac{3}{\sqrt{7-2\sqrt{10} } } = \dfrac{3}{\sqrt{5-2\sqrt{5}\sqrt{2} + 2} } = \dfrac{3}{\sqrt{(\sqrt{5})^{2}-2\sqrt{5}\sqrt{2} + (\sqrt{2})^{2}} }=\\ \\ \\ \\ \dfrac{3}{\sqrt{(\sqrt{5}-\sqrt{2})^{2}}} }= \dfrac{3}{\sqrt{5}-\sqrt{2}}=\dfrac{3}{\sqrt{5}-\sqrt{2}}\cdot \dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\\ \\ \\ \\ \dfrac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}=\dfrac{3(\sqrt{5}+\sqrt{2})}{5-2}=\dfrac{3(\sqrt{5}+\sqrt{2})}{3}=\boxed{\sqrt{5}+\sqrt{2}}[/tex]

Resolvendo a segunda fração:

[tex]\dfrac{4}{\sqrt{8+4\sqrt{3} } }=\dfrac{4}{\sqrt{6+4\sqrt{3}+2 } } =[/tex]

Observe que:

[tex]\sf 4\sqrt{3}=\sqrt{16\cdot 3}=\sqrt{4\cdot 4\cdot 3}=\sqrt{2^{2}\cdot 2^{2}\cdot 3}=\\ \\ \\ 2\sqrt{2^{2}\cdot 3}=2\sqrt{2\cdot 2\cdot 3}=2\sqrt{2}\sqrt{2}\sqrt{3}[/tex]

Continuando com a construção do quadrado:

[tex]\dfrac{4}{\sqrt{2\cdot 3+2\sqrt{2}\sqrt{2}\sqrt{3}+2 } } = \dfrac{4}{\sqrt{(\sqrt{2})^{2}\cdot (\sqrt{3})^{2}+2\sqrt{2}\sqrt{2}\sqrt{3}+(\sqrt{2})^{2} } } =\\ \\ \\ \\ \dfrac{4}{\sqrt{\left(\sqrt{2}\cdot \sqrt{3}\right)^{2}+2\sqrt{2}\sqrt{2}\sqrt{3}+(\sqrt{2})^{2} } } = \dfrac{4}{\sqrt{\left(\sqrt{2}\cdot \sqrt{3} + \sqrt{2}\right)^{2}}}=[/tex]

[tex]\dfrac{4}{\sqrt{2}\cdot \sqrt{3} + \sqrt{2}}=\dfrac{4}{\sqrt{6} + \sqrt{2}}= \dfrac{4}{\sqrt{6} + \sqrt{2}}\cdot \dfrac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}=\\ \\ \\ \\ \dfrac{4(\sqrt{6} - \sqrt{2})}{(\sqrt{6})^{2} - (\sqrt{2})^{2}}=\dfrac{4(\sqrt{6} - \sqrt{2})}{6 - 2}=\dfrac{4(\sqrt{6} - \sqrt{2})}{4}= \boxed{\sqrt{6} - \sqrt{2}}[/tex]

Resolvendo a terceira fração:

[tex]\dfrac{1}{\sqrt{11-2\sqrt{30} } } = \dfrac{1}{\sqrt{6-2\sqrt{6}\sqrt{5} + 5} } =\dfrac{1}{\sqrt{(\sqrt{6})^{2}-2\sqrt{6}\sqrt{5} + (\sqrt{5})^{2}} } \\ \\ \\ \\ \dfrac{1}{\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^{2}} }= \dfrac{1}{\sqrt{6}-\sqrt{5}}= \dfrac{1}{\sqrt{6}-\sqrt{5}}\cdot \dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\\ \\ \\ \\ \dfrac{\sqrt{6}+\sqrt{5}}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}= \dfrac{\sqrt{6}+\sqrt{5}}{6-5}=\boxed{\sqrt{6}+\sqrt{5}}[/tex]

Agora, pra concluir:

[tex]\sf \sqrt{5}+\sqrt{2}+\sqrt{6} - \sqrt{2}=\sqrt{6}+\sqrt{5}\\ \\ \\ \\ \sqrt{5}+\sqrt{6} =\sqrt{6}+\sqrt{5}[/tex]