\(\left(\sqrt{12}+2\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)

\(=\sqrt{12}:\sqrt{3}+2\sqrt{27}:\sqrt{3}-\sqrt{3}:\sqrt{3}\)

\(=\sqrt{4}+2\sqrt{9}-1\)

\(=2+6-1\)

\(=7\)

2) \(\left(4\sqrt{2}-\sqrt{8}+2\right).\sqrt{2-\sqrt{8}}\)

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

\(=\left(2\sqrt{2}+2\right)^2.\left(\sqrt{2-2\sqrt{2}}\right)^2\)

\(=\left(8+4\right)\left(2-2\sqrt{2}\right)\)

\(=12.\left(2-2\sqrt{2}\right)\)

\(=24-24\sqrt{2}\)

\(=24\left(1-\sqrt{2}\right)\)

3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\frac{3}{2}\sqrt{12}\right)\)

\(=\sqrt{3}\left(2\sqrt{3^2.3}-\sqrt{5^2.3}+\frac{3}{2}\sqrt{2^2.3}\right)\)

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

\(=\sqrt{3}.4\sqrt{3}\)

\(=12\)

1.\(D=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10\sqrt{3}}{3}\)\(=\frac{-17\sqrt{3}}{3}\)

2.\(A=27-\left(\sqrt{32}-\sqrt{50}\right)^2=25\)

\(B=1-\left(\left(-\sqrt{3}\right)^2-\left(\sqrt{20}-\sqrt{45}\right)^2\right)\)\(=1-\left(-2\right)=3\)

\(a,=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\\ b,=\left|1-\sqrt{2}\right|+\sqrt{5}=\sqrt{2}-1+\sqrt{5}\)

a)\(\sqrt{12}+3\sqrt{27}-\sqrt{300}=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\)

b) \(\sqrt{\left(1-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+3\right)^2}=\left|1-\sqrt{2}\right|-\left|\sqrt{2}+3\right|=\sqrt{2}-1-\sqrt{2}-3=-4\)

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+1\right)=-5-10\sqrt{6}\)        

Mình không chắc lắm

\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)

\(=9\sqrt{5}-28\sqrt{3}\)

\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)

\(=4-2\cdot5+6-7\)

\(=4-10+6-7\)

=-7

A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)

  =\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)

  =2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)

   =\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)

   =9\(\sqrt{5}\)- 28\(\sqrt{3}\)