Bài 2:
a: Ta có: \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)
\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)
\(=-\sqrt{3}\)
b: ta có: \(\sqrt{98}-\sqrt{72}+0.5\sqrt{8}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}\)
\(=2\sqrt{2}\)
c: ta có: \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)
\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
\(=6\sqrt{a}\)
d: Ta có: \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)
\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
\(=4\sqrt{b}-5\sqrt{10b}\)
Bài 3:
a: Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)
\(=6+\sqrt{15}-2\sqrt{15}\)
\(=6-\sqrt{15}\)
b: Ta có: \(\left(5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}+10-5\sqrt{10}\)
=10
c: ta có: \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
=7
d: Ta có: \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}\)
\(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}\)
=22
