Bài 4:
a: \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{6}}{3}+\dfrac{5\sqrt{6}}{3}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\)
b: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
c: \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\sqrt{3}+\dfrac{10}{\sqrt{3}}\right)\cdot2\sqrt{3}\)
\(=\left(-\dfrac{1}{2}\sqrt{3}+\dfrac{10}{\sqrt{3}}\right)\cdot2\sqrt{3}\)
\(=-\dfrac{1}{2}\sqrt{3}\cdot2\sqrt{3}+\dfrac{10}{\sqrt{3}}\cdot2\sqrt{3}\)
\(=-3+20=17\)
d: \(\sqrt{3-\sqrt{5}}\cdot\sqrt{8}=2\cdot\sqrt{6-2\sqrt{5}}\)
\(=2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\left(\sqrt{5}-1\right)=2\sqrt{5}-2\)
