[tex]\it \sqrt{99}\cdot\sqrt{2,(27)}\cdot\sqrt{1,(3)}\cdot\sqrt{\dfrac{1}{3}}=\sqrt{9\cdot11}\cdot\sqrt{2\dfrac{27}{99}}\cdot\sqrt{1\dfrac{3}{9}}\cdot\sqrt{\dfrac{1}{3}}=\\ \\ \\ =3\sqrt{11\cdot\dfrac{225}{99}\cdot\dfrac{\not12\ ^4}{9}\cdot\dfrac{1}{\not3}}=3\sqrt{\dfrac{25\cdot4}{9}}=3\cdot\dfrac{5\cdot2}{3}=10[/tex]
