a, Ta có : \(A=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
=> \(A=3-\sqrt{5}+\sqrt{5}-2=1\)
b, Ta có : \(B=\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
=> \(B=\sqrt{2}\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
=> \(B=\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}\)
=> \(B=\left(\sqrt{5}+1\right)\left(\sqrt{6-2\sqrt{5}}\right)\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
=> \(B=\left(\sqrt{5}+1\right)\left(\sqrt{5-2\sqrt{5}+1}\right)\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
=> \(B=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
=> \(B=\left(5-1\right)\sqrt{36-20}=4.\sqrt{16}=4.4=16\)