a/ ĐKXĐ: \(\left\{{}\begin{matrix}1+x\ge0\\1-x\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le1\end{matrix}\right.\Leftrightarrow-1\le x\le1\)
b/ \(A=\left(\frac{1}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\frac{1}{\sqrt{1-x^2}}+1\right)\)
\(=\frac{1+\sqrt{1+x}.\sqrt{1-x}}{\sqrt{1+x}}:\frac{1+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)
\(=\frac{1+\sqrt{1-x^2}}{\sqrt{1+x}}.\frac{\sqrt{1-x^2}}{1+\sqrt{1-x^2}}\)
\(=\sqrt{1-x}\)
c/ Ta có: \(x=\frac{\sqrt{3}}{2+\sqrt{3}}=\frac{\sqrt{3}.\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{2\sqrt{3}-3}{4-3}=2\sqrt{3}-3\)
Vậy \(A=\sqrt{1-\left(2\sqrt{3}-3\right)}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)