Question

giúp mình vs ạ !

Answer 1

a: \(5\sqrt{27}+3\sqrt{48}-2\cdot\sqrt{12}-6\sqrt3\)

\(=5\cdot3\sqrt3+3\cdot4\sqrt3-2\cdot2\sqrt3-6\sqrt3\)

\(=15\sqrt3+12\sqrt3-4\sqrt3-6\sqrt3=17\sqrt3\)

b: \(\left(3+\frac{3-\sqrt3}{1-\sqrt3}\right)\left(\frac{\sqrt{21}+\sqrt7}{\sqrt7}+2\right)\)
\(=\left(3-\frac{\sqrt3\left(\sqrt3-1\right)}{\sqrt3-1}\right)\left(\frac{\sqrt7\left(\sqrt3+1\right)}{\sqrt7}+2\right)\)

\(=\left(3-\sqrt3\right)\left(\sqrt3+1+2\right)=\left(3-\sqrt3\right)\left(3+\sqrt3\right)\)

=9-3

=6

c: \(\sqrt{96}-6\cdot\sqrt{\frac23}+\frac{3}{3+\sqrt6}-\sqrt{10-4\sqrt6}\)

\(=4\sqrt6-6\cdot\frac{\sqrt6}{3}+\frac{3\left(3-\sqrt6\right)}{\left(3+\sqrt6\right)\left(3-\sqrt6\right)}-\sqrt{\left(\sqrt6-2\right)^2}\)

\(=4\sqrt6-2\sqrt6+\frac{3\left(3-\sqrt6\right)}{9-6}-\left(\sqrt6-2\right)=2\sqrt6-\sqrt6+2+3-\sqrt6=5\)

d: \(\sqrt{\frac52-\sqrt6}-\sqrt{\frac{11}{2}-2\sqrt6}\)

\(=\sqrt{\frac{5-2\sqrt6}{2}}-\sqrt{\frac{11-4\sqrt6}{2}}=\sqrt{\frac{\left(\sqrt3-\sqrt2\right)^2}{2}}-\sqrt{\frac{\left(2\sqrt2-\sqrt3\right)^2}{2}}\)

\(=\frac{\sqrt3-\sqrt2-\left(2\sqrt2-\sqrt3\right)}{\sqrt2}=\frac{\sqrt3-\sqrt2-2\sqrt2+\sqrt3}{\sqrt2}=\frac{2\sqrt3-3\sqrt2}{\sqrt2}=\frac{2\sqrt6-6}{2}\)