Ta có:
\(\sqrt{\dfrac{7+3\sqrt{5}}{2}}\)
\(=\sqrt{\dfrac{2\cdot\left(7+3\sqrt{5}\right)}{2\cdot2}}\)
\(=\sqrt{\dfrac{14+6\sqrt{5}}{4}}\)
\(=\sqrt{\dfrac{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot3-3^2}{2^2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{5}+3\right)^2}{2^2}}\)
\(=\dfrac{3+\sqrt{5}}{2}\)
Mà: \(\dfrac{3+\sqrt{5}}{2}=a+b\sqrt{5}\)
Nên: \(\dfrac{3+\sqrt{5}}{2}=\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}=\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}\)
Vậy: \(a=\dfrac{3}{2};b=\dfrac{1}{2}\)
\(\Rightarrow a+b=\dfrac{3}{2}+\dfrac{1}{2}=\dfrac{4}{2}=2\)
\(\sqrt{\dfrac{7+3\sqrt{5}}{2}}=\sqrt{\dfrac{14+6\sqrt{5}}{4}}=\sqrt{\left(\dfrac{3+\sqrt{5}}{2}\right)^2}\)
\(=\dfrac{3+\sqrt{5}}{2}\)
=>a=3/2; b=1/2
a+b=3/2+1/2=2
