Lời giải:
Đặt \(\sqrt{1+\frac{\sqrt{3}}{4}}=a; \sqrt{1-\frac{\sqrt{3}}{4}}=b\) $(a,b\geq 0)$
Khi đó:
$a^2+b^2=2$
$ab=\sqrt{1-\frac{3}{16}}=\frac{\sqrt{13}}{4}$
$a+b=\sqrt{(a+b)^2}=\sqrt{a^2+b^2+2ab}=\sqrt{2+\frac{\sqrt{13}}{2}}$
\(=\sqrt{\dfrac{4+\sqrt{3}}{4}}+\sqrt{1-\dfrac{\sqrt{3}}{4}}\)
\(=\dfrac{\sqrt{4+\sqrt{3}}}{\sqrt{4}}+\sqrt{1-\dfrac{\sqrt{3}}{4}}\)
\(=\dfrac{\sqrt{4+\sqrt{3}}}{2}+\sqrt{1-\dfrac{\sqrt{3}}{4}}\)
\(=\dfrac{\sqrt{4+\sqrt{3}}}{2}+\sqrt{\dfrac{4-\sqrt{3}}{4}}\)
\(=\dfrac{\sqrt{4+\sqrt{3}}}{2}+\dfrac{\sqrt{4-\sqrt{3}}}{2}\)
\(=\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{2}\)
