Question

Transforme os radicais duplos na soma ou diferença de dois radicais simples. PFVR, ME AJUDEM!!!
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Answer 1

Explicação passo-a-passo:

$\sqrt{A+\sqrt{B}}=\sqrt{\dfrac{A+C}{2}}+\sqrt{\dfrac{A-C}{2}}$, sendo $C=\sqrt{A^2-B}$

a) $\sqrt{7+\sqrt{40}}$

$C=\sqrt{7^2-40}$

$C=\sqrt{49-40}$

$C=\sqrt{9}$

$C=3$

$\sqrt{7+\sqrt{40}}=\sqrt{\dfrac{7+3}{2}}+\sqrt{\dfrac{7-3}{2}}$

$\sqrt{7+\sqrt{40}}=\sqrt{\dfrac{10}{2}}+\sqrt{\dfrac{4}{2}}$

$\sqrt{7+\sqrt{40}}=\sqrt{5}+\sqrt{2}$

b) $\sqrt{13+2\sqrt{30}}=\sqrt{13+\sqrt{2^2\cdot30}}=\sqrt{13+\sqrt{120}}$

$C=\sqrt{13^2-120}$

$C=\sqrt{169-120}$

$C=\sqrt{49}$

$C=7$

$\sqrt{13+2\sqrt{30}}=\sqrt{\dfrac{13+7}{2}}+\sqrt{\dfrac{13-3}{2}}$

$\sqrt{13+2\sqrt{30}}=\sqrt{\dfrac{20}{2}}+\sqrt{\dfrac{10}{2}}$

$\sqrt{13+2\sqrt{30}}=\sqrt{10}+\sqrt{5}$

c) $\sqrt{\sqrt{7+4\sqrt{3}}}=\sqrt{\sqrt{7+\sqrt{4^2\cdot3}}}=\sqrt{\sqrt{7+\sqrt{48}}}$

$C=\sqrt{7^2-48}$

$C=\sqrt{49-48}$

$C=\sqrt{1}$

$C=1$

$\sqrt{\sqrt{7+4\sqrt{3}}}=\sqrt{\sqrt{\dfrac{7+1}{2}}+\sqrt{\dfrac{7-1}{2}}}$

$\sqrt{\sqrt{7+4\sqrt{3}}}=\sqrt{\sqrt{\dfrac{8}{2}}+\sqrt{\dfrac{6}{2}}}$

$\sqrt{\sqrt{7+4\sqrt{3}}}=\sqrt{\sqrt{4}+\sqrt{3}}$

$\sqrt{\sqrt{7+4\sqrt{3}}}=\sqrt{2+\sqrt{3}}$

$C=\sqrt{2^2-3}$

$C=\sqrt{4-3}$

$C=\sqrt{1}$

$C=1$

$\sqrt{2+\sqrt{3}}=\sqrt{\dfrac{2+1}{2}}+\sqrt{\dfrac{2-1}{2}}$

$\sqrt{2+\sqrt{3}}=\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{1}{2}}$

$\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3}}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}$

$\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2}$

Logo:

$\sqrt{\sqrt{7+4\sqrt{3}}}=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2}$

d) $\sqrt{A-\sqrt{B}}=\sqrt{\dfrac{A+C}{2}}-\sqrt{\dfrac{A-C}{2}}$, sendo $C=\sqrt{A^2-B}$

$\sqrt{6-2\sqrt{5}}=\sqrt{6-\sqrt{2^2\cdot5}}=\sqrt{6-\sqrt{20}}$

$C=\sqrt{6^2-20}$

$C=\sqrt{36-20}$

$C=\sqrt{16}$

$C=4$

$\sqrt{6-2\sqrt{5}}=\sqrt{\dfrac{6+4}{2}}-\sqrt{\dfrac{6-4}{2}}$

$\sqrt{6-2\sqrt{5}}=\sqrt{\dfrac{10}{2}}-\sqrt{\dfrac{2}{2}}$

$\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{1}$

$\sqrt{6-2\sqrt{5}}=\sqrt{5}-1$