Bài 1:
a) \(\dfrac{12}{3-\sqrt{3}}=\dfrac{12\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\dfrac{36+12\sqrt{3}}{6}\)
b) \(\dfrac{8}{\sqrt{5}+2}=\dfrac{8\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}=8\sqrt{5}-16\)
c) \(\dfrac{14}{\sqrt{10}+\sqrt{3}}=\dfrac{14\left(\sqrt{10}-\sqrt{3}\right)}{\left(\sqrt{10}+\sqrt{3}\right)\left(\sqrt{10}-\sqrt{3}\right)}=\dfrac{14\left(\sqrt{10}-\sqrt{3}\right)}{7}=2\sqrt{10}-2\sqrt{3}\)
Bài 2:
a: Ta có: \(\sqrt{75}-\sqrt{300}+\sqrt{48}\)
\(=5\sqrt{3}-10\sqrt{3}+4\sqrt{3}\)
\(=-\sqrt{3}\)
b: Ta có: \(\sqrt{98}+\dfrac{1}{2}\sqrt{8}-\sqrt{72}\)
\(=7\sqrt{2}+\dfrac{1}{2}\cdot2\sqrt{2}-6\sqrt{2}\)
\(=2\sqrt{2}\)
c: Ta có: \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=3\sqrt{2}+8\sqrt{2}-3\sqrt{2}\)
\(=8\sqrt{2}\)
d: Ta có: \(\sqrt{3}-\dfrac{1}{3}\sqrt{27}+2\sqrt{507}\)
\(=\sqrt{3}-\dfrac{1}{3}\cdot3\sqrt{3}+26\sqrt{3}\)
\(=26\sqrt{3}\)
e: Ta có: \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)
\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{b}\)
\(=6\sqrt{a}\)
Bài 1:
a: \(\dfrac{12}{3-\sqrt{3}}=6+2\sqrt{3}\)
b: \(\dfrac{8}{\sqrt{5}+2}=-16+8\sqrt{5}\)
c: \(\dfrac{14}{\sqrt{10}+\sqrt{3}}=2\sqrt{10}-2\sqrt{3}\)
e: \(\dfrac{3\sqrt{5}-2\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}=\dfrac{18+5\sqrt{10}}{2}\)
