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\(a.\sqrt{200}-\sqrt{32}+\sqrt{72}=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
\(b.\sqrt{175}-\sqrt{112}+\sqrt{63}=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
\(c.4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
a)\(2\sqrt{18}+3\sqrt{8}-3\sqrt{32}-\sqrt{50}\)
\(=2\sqrt{9.2}+3\sqrt{4.2}-3\sqrt{16.2}-\sqrt{25.2}\)
\(=6\sqrt{2}+6\sqrt{2}-12\sqrt{2}-5\sqrt{2}\)
\(=-5\sqrt{2}\)
b) \(\sqrt{200}-\sqrt{32}-\sqrt{72}\)
\(=\sqrt{100.2}-\sqrt{16.2}-\sqrt{36.2}\)
\(=10\sqrt{2}-4\sqrt{2}-6\sqrt{2}\)
\(=0\)
c) \(\sqrt{175}-\sqrt{112}+\sqrt{63}\)
\(=\sqrt{25.7}-\sqrt{16.7}+\sqrt{9.7}\)
\(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}\)
\(=4\sqrt{7}\)
d) \(3\sqrt{8}-\sqrt{32}+4\sqrt{2}+\sqrt{162}\)
\(=3\sqrt{4.2}-\sqrt{16.2}+4\sqrt{2}+\sqrt{81.2}\)
\(=6\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}\)
\(=15\sqrt{2}\)
a: Ta có: \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)
\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}-2\sqrt{5}\)
\(=-\sqrt{7}-2\sqrt{5}\)
\(-\dfrac{6\sqrt{2}-\sqrt{\left(9-8\sqrt{2}\right)\cdot2}}{2}\)
\(\sqrt{4\dfrac{1}{2}}+\sqrt{32}-\sqrt{72}+\sqrt{162}\\ =\sqrt{\dfrac{4\cdot2+1}{2}}+\sqrt{4^2\cdot2}-\sqrt{6^2\cdot2}+\sqrt{9^2\cdot2}\\ =\sqrt{\dfrac{9}{2}}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\\ =\dfrac{3}{\sqrt{2}}+7\sqrt{2}\\ =\dfrac{3}{\sqrt{2}}+\dfrac{7\sqrt{2}\cdot\sqrt{2}}{\sqrt{2}}\\ =\dfrac{17}{\sqrt{2}}\)
\(=\sqrt{\dfrac{9}{2}}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\)
\(=\dfrac{3}{2}\sqrt{2}+7\sqrt{2}=\dfrac{17}{2}\sqrt{2}\)
6: Ta có: \(\left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)\)
=18-3
=15
7: Ta có: \(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)
\(=6\sqrt{2}+\dfrac{3}{2}\sqrt{2}-4\sqrt{2}-9\sqrt{2}\)
\(=-\dfrac{11}{2}\sqrt{2}\)
Câu a : \(\sqrt{200}-\sqrt{32}+\sqrt{72}-\sqrt{162}\)
\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)
\(=3\sqrt{2}\)
Câu b : \(\sqrt{63}+\sqrt{175}+\sqrt{112}\)
\(=3\sqrt{7}+5\sqrt{7}+4\sqrt{7}\)
\(=12\sqrt{7}\)
Học tốt !!!!!!!!!!!!!