Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=9\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}=18\sqrt{2}\)
b: \(=8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)
c: \(=2\sqrt{21}\)
a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)
\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)
\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)
\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)
\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)
\(=-8\sqrt{3}\)
b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)
\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)
6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)
7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)
8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)
9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)
1) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=-6\sqrt{2}\)
2) \(\sqrt{50}-\sqrt{18}+\sqrt{200}-\sqrt{162}\)
\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}\)
\(=3\sqrt{2}\)
3) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
\(=5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
\(=-2\sqrt{5}\)
4) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=4\sqrt{3}\)
5) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}\)
\(=-\dfrac{17}{3}\sqrt{3}\)
a) Ta có: \(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{12}{3-\sqrt{3}}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{1}+\dfrac{12\left(3+\sqrt{3}\right)}{6}\)
\(=\sqrt{3}+1-6-3\sqrt{3}+6+2\sqrt{3}\)
\(=1\)
b) Ta có: \(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\sqrt{7}+\sqrt{5}}-\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{2}+\sqrt{7}-\sqrt{3}\)
=0