a/ Đề sai

b/ \(\sqrt{125}-4\sqrt{45}+3\sqrt{2}-\sqrt{80}=5\sqrt{5}-12\sqrt{5}+3\sqrt{2}-4\sqrt{5}\)

\(=-11\sqrt{5}+3\sqrt{2}\)

c/ \(2\sqrt{\frac{27}{4}}-\sqrt{\frac{48}{9}}-\frac{2}{5}\sqrt{\frac{75}{16}}=2.\frac{3\sqrt{3}}{2}-\frac{4\sqrt{3}}{3}-\frac{2}{5}.\frac{5\sqrt{3}}{4}\)

\(=3\sqrt{3}-\frac{4\sqrt{3}}{3}-\frac{\sqrt{3}}{2}=\sqrt{3}\left(3-\frac{4}{3}-\frac{1}{2}\right)=\frac{7\sqrt{3}}{6}\)

d/ \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}=33-3\sqrt{22}-11+3\sqrt{22}=22\)

a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)

\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)

\(=3\sqrt{2}-2\sqrt{3}\)

b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)

\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)

\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)

c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)

\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)

\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)

\(=75-20=55\)

d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)

\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)

\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)

\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)

\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)

a) \(\sqrt{36}.\sqrt{121}+\sqrt[3]{-64}-\sqrt[3]{125}\)

\(=6.11+\left(-4\right)-5=66-9=57\)

b) \(\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}-30\sqrt{\frac{3}{25}}\)

\(=\sqrt{25.3}+\left|\sqrt{3}-2\right|-30.\frac{\sqrt{3}}{\sqrt{25}}\)

\(=5\sqrt{3}+2-\sqrt{3}-30.\frac{\sqrt{3}}{5}\)

\(=5\sqrt{3}+2-\sqrt{3}-6\sqrt{3}=2-2\sqrt{3}\)

c) \(\sqrt{11-4\sqrt{7}}-\frac{12}{1+\sqrt{7}}=\sqrt{7-4\sqrt{7}+4}-\frac{12}{1+\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\frac{12}{1+\sqrt{7}}=\left|\sqrt{7}-2\right|-\frac{12}{1+\sqrt{7}}\)

\(=\left(\sqrt{7}-2\right)-\frac{12}{\sqrt{7}+1}=\frac{\left(\sqrt{7}-2\right)\left(\sqrt{7}+1\right)}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}\)

\(=\frac{5-\sqrt{7}}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}=\frac{-7-\sqrt{7}}{\sqrt{7}+1}\)

\(=\frac{-\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}=-\sqrt{7}\)

a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)

\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)

\(=11\sqrt{2}\)

b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)

\(=5\sqrt{5}-1\)

a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)

\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)

\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)

\(=\left(10-4+6\right)\sqrt{2}\)

\(=12\sqrt{2}\)

b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)

\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)

\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)

\(=\left(8-15+15-3\right)\sqrt{5}\)

\(=5\sqrt{5}\)

c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)

\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)

\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)

1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)

2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)

3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)

4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)

5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)

a: Ta có: \(\sqrt{75}-2\sqrt{27}+\sqrt{48}\)

\(=5\sqrt{3}-2\cdot3\sqrt{3}+4\sqrt{3}\)

\(=3\sqrt{3}\)

c: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)

\(=\sqrt{7}+1-\sqrt{7}+2\)

=3