Biểu thức \(\sqrt 3 \cdot \sqrt {16} \cdot \sqrt {14} \) bằng

Chọn A

Ta có \[\left( {1 + \sqrt {\frac{3}{5}} } \right)\left( {1 - \sqrt {\frac{3}{5}} } \right)\]\[ = {1^2} - {\left( {\sqrt {\frac{3}{5}} } \right)^2}\]\( = 1 - \frac{3}{5}\)\( = \frac{2}{5}\).

Suy ra \(a = 2\), \(b = 5\).

Vậy \(ab = 2.5 = 10\).

Chọn B

Ta có \[ - \frac{1}{3}a{b^3} \cdot \sqrt {\frac{{9{a^2}}}{{{b^6}}}} \]\[ =  - \frac{1}{3}a{b^3} \cdot \frac{{\sqrt {9{a^2}} }}{{\sqrt {{b^6}} }}\]

\[ =  - \frac{1}{3}a{b^3} \cdot \frac{{\sqrt {{{\left( {3a} \right)}^2}} }}{{\sqrt {{{\left( {{b^3}} \right)}^2}} }}\]\[ =  - \frac{1}{3}a{b^3} \cdot \frac{{\left| {3a} \right|}}{{\left| {{b^3}} \right|}}\]

\[ =  - \frac{1}{3}a{b^3} \cdot \frac{{ - 3a}}{{{b^3}}}\]\( = {a^2}\).