\(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\\ A=\sqrt{8-2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\\ A=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\\ A=\left(\sqrt{5}-\sqrt{3}\right)^2\left(4+\sqrt{15}\right)\\ A=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\\ A=2\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)=2\left[4^2-\left(\sqrt{15}\right)^2\right]=2\cdot1=2\)
\(A=\sqrt{4-\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=2>\sqrt{3}\)
xét vế trái ta có
\(A=\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)
A=\(\sqrt{4^2-\sqrt{15}^2}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)
A=\(\sqrt{4+\sqrt{15}}.\sqrt{\sqrt{10}-\sqrt{6}}.\sqrt{10-\sqrt{6}}\)
A=\(\sqrt{\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)}.\sqrt{\sqrt{10}-\sqrt{6}}\)
A=\(\sqrt{4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}}.\sqrt{\sqrt{10}-\sqrt{6}}\)
A=\(\sqrt{4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}}.\sqrt{\sqrt{10}-\sqrt{6}}\)
A=\(\sqrt{\sqrt{10}+\sqrt{6}}.\sqrt{\sqrt{10}-\sqrt{6}}\)
A=\(\sqrt{\sqrt{10}^2-\sqrt{6}^2}=\sqrt{4}\)
mà:\(\sqrt{4}>\sqrt{3}\) nên A\(>\sqrt{3}\)
