a) Ta có: \(2\sqrt{3}=\sqrt{4\cdot3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{9\cdot2}=\sqrt{18}\)
mà \(\sqrt{12}< \sqrt{18}\)(Vì 12<18)
nên \(2\sqrt{3}< 3\sqrt{2}\)
b) Ta có: \(A=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|+\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2+\sqrt{5}+2\)(Vì \(\sqrt{5}>2>0\))
\(=2\sqrt{5}\)