\(A=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}\)
\(A^2=8+2\sqrt{15}+2\sqrt{\left(8+2\sqrt{15}\right)\left(8-2\sqrt{15}\right)}+8-2\sqrt{15}\)
\(A^2=16+2\sqrt{64-15}\)
\(A^2=16+2.50\)
\(A^2=116\)
Vậy \(\orbr{\begin{cases}A=\sqrt{116}\\A=-\sqrt{116}\end{cases}}\)
Chúc bạn học tốt ~
1) \(\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{5}+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}>\sqrt{3}\) )
\(=2\sqrt{5}\)
2) \(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{2}+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{2}+\sqrt{3}+\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}>\sqrt{3}\) )
\(=\sqrt{2}+\sqrt{5}\)
