\(\left(\sqrt{1-x}.\dfrac{\sqrt{3}}{\sqrt{1-x}}+\sqrt{x}.\dfrac{2}{\sqrt{x}}\right)^2\le\left(1-x+x\right)\left(B\right)\)
\(\Rightarrow B\ge\left(\sqrt{3}+2\right)^2=7+4\sqrt{3}\)
Bmin = 7+4can 3
khi\(\dfrac{\sqrt{3}}{1-x}=\dfrac{2}{x}\Rightarrow x=\dfrac{2}{\sqrt{3}+2}\)