\(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\)
= \(\sqrt{2.2^2}-2\sqrt{4^2.2}+3\sqrt{5^2.2}\)
= \(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\)
= \(9\sqrt{2}\)
\(\dfrac{1}{3}+\sqrt{2}-\dfrac{1}{3}-\sqrt{2}\)
= \(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\left(\sqrt{2}-\sqrt{2}\right)\)
= 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
