\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(A=\sqrt{8-2\cdot\sqrt{3}\cdot\sqrt{5}}-\sqrt{8+2\cdot\sqrt{3}\cdot\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(A=\left|\sqrt{5}-\sqrt{3}\right|-\left|\sqrt{5}+\sqrt{3}\right|\)
\(A=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(A=-2\sqrt{3}\)
Cách khác:
\(A^2=\left(\sqrt{8-2\sqrt{15}}\right)^2-2.\sqrt{8-2\sqrt{15}}.\sqrt{8+2\sqrt{15}}+\left(\sqrt{8+2\sqrt{15}}\right)^2\)
\(A^2=8-2\sqrt{15}-2.\sqrt{8^2-\left(2\sqrt{15}\right)^2}+8+2\sqrt{15}\)
\(A^2=16-2.2=12\)\(\Rightarrow\left[{}\begin{matrix}A=2\sqrt{3}\\A=-2\sqrt{3}\end{matrix}\right.\)
Vì \(\sqrt{8-2\sqrt{15}}< \sqrt{8+2\sqrt{15}}\) nên A<0 nên A=\(-2\sqrt{3}\)