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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: \(=2\sqrt{21}\)
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: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)
a, (\(\sqrt{128}\)-\(\sqrt{50}\)+\(\sqrt{98}\)):\(\sqrt{2}\)
=(8-5+3)
=10
b, (\(\sqrt{48}\)+\(\sqrt{27}\)-\(\sqrt{192}\)):2\(\sqrt{3}\)
=(2+1,5-4)
=-0,5
c, \(\dfrac{1}{8}\)-3\(\sqrt{2}\) +\(\dfrac{1}{8}\)+3\(\sqrt{2}\)
=\(\dfrac{1}{4}\)
d, \(\sqrt{\left(1-\sqrt{5}\right)^2}-\sqrt{5}\)
=-1
\(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\)
= \(\sqrt{2.2^2}-2\sqrt{4^2.2}+3\sqrt{5^2.2}\)
= \(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\)
= \(9\sqrt{2}\)
\(\dfrac{1}{3}+\sqrt{2}-\dfrac{1}{3}-\sqrt{2}\)
= \(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\left(\sqrt{2}-\sqrt{2}\right)\)
= 0
\(\left(\sqrt{12}+\sqrt{27}-\sqrt{108}\right).2\sqrt{3}\)
= \(\left(2\sqrt{3}+3\sqrt{3}-6\sqrt{3}\right).2\sqrt{3}\)
= \(-\sqrt{3}.2\sqrt{3}\)
= \(-6\)
b:Ta có: \(\dfrac{1}{3+\sqrt{2}}-\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{3-\sqrt{2}-3-\sqrt{2}}{7}\)
\(=\dfrac{-2\sqrt{2}}{7}\)
a: Ta có: \(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\)
\(=2\sqrt{2}-2\cdot4\sqrt{2}+3\cdot5\sqrt{2}\)
\(=9\sqrt{2}\)
c: Ta có: \(\left(\sqrt{12}+\sqrt{27}-\sqrt{108}\right)\cdot2\sqrt{3}\)
\(=\left(2\sqrt{3}+3\sqrt{3}-6\sqrt{3}\right)\cdot2\sqrt{3}\)
=-6