Bài 1: Rút gọn
a) Ta có: \(\left(\sqrt{3}-\sqrt{2}+1\right)\cdot\left(\sqrt{3}-1\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-1\right)-\sqrt{2}\cdot\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}\)
\(=2-\sqrt{2}-\sqrt{6}\)
b) Ta có: \(0.2\cdot\sqrt{\left(-10\right)^2\cdot3}+2\cdot\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
\(=0.2\cdot\sqrt{\left(-10\right)^2}\cdot\sqrt{3}+2\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=0.2\cdot10\cdot\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{5}\)
c) Ta có: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\sqrt{196}-2\cdot\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-\sqrt{392}+7+\sqrt{392}\)
=21
d) Ta có: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=\sqrt{5}\left(15+5\cdot2-3\cdot3\right)\)
\(=16\sqrt{5}\)
a, \(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)-\sqrt{2}\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}=2+\sqrt{2}-\sqrt{6}\)
b, \(=\sqrt{300.0,04}+2\left|\sqrt{3}-\sqrt{5}\right|\)
\(=2\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c, \(=\sqrt{196}-2\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
d, \(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+10\sqrt{5}-9\sqrt{5}=16\sqrt{5}\)