a. P=\(\left(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{3}{\sqrt{1-a^2}}+1\right)=\left(\dfrac{3}{\sqrt{1+a}}+\dfrac{\sqrt{1-a^2}}{\sqrt{1+a}}\right):\left(\dfrac{3}{\sqrt{1-a^2}}+\dfrac{\sqrt{1-a^2}}{\sqrt{1-a^2}}\right)=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}:\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3+\sqrt{1-a^2}}=\sqrt{1-a}\)
b. Thay \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}\) vào P, ta có:
\(P=\sqrt{1-\dfrac{\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\dfrac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\dfrac{2}{2+\sqrt{3}}}=\dfrac{\sqrt{2}}{\sqrt{2+\sqrt{3}}}=\dfrac{\sqrt{2}}{\sqrt{\left(\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}}=\dfrac{\sqrt{2}}{\dfrac{\sqrt{3}+1}{\sqrt{2}}}=\dfrac{2}{\sqrt{3}+1}=\dfrac{2\left(\sqrt{3}-1\right)}{3-1}=-1+\sqrt{3}\)
Vậy giá trị của P tại \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}\) là \(-1+\sqrt{3}\)
