Question

\(\sqrt{7+\sqrt{13}}\) + \(\sqrt{7-\sqrt{13}}\)

Answer 1

Đặt \(A=\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}\Rightarrow A^2=7+\sqrt{13}+7-\sqrt{13}+2\sqrt{\left(7+\sqrt{13}\right)\left(7-\sqrt{13}\right)}=14+2\sqrt{49-13}=14+2\sqrt{36}=14+12=26\Rightarrow A=\pm\sqrt{26}\)Mà \(\left\{{}\begin{matrix}\sqrt{7+\sqrt{13}}>0\\\sqrt{7-\sqrt{13}}>0\end{matrix}\right.\)⇒\(\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}>0\Rightarrow A>0\)

Vậy \(A=\sqrt{26}\Rightarrow\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}=\sqrt{26}\)

Answer 2

Cách khác :

\(\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}=\dfrac{\sqrt{14+2\sqrt{13}}+\sqrt{14-2\sqrt{13}}}{\sqrt{2}}=\dfrac{\sqrt{13+2\sqrt{13}+1}+\sqrt{13-2\sqrt{13}+1}}{\sqrt{2}}=\dfrac{\sqrt{13}+1+\sqrt{13}-1}{\sqrt{2}}=\sqrt{13}\)