Tính a + b .

Do \(\frac{\pi }{2} < \alpha ,\beta < \pi \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\cos \alpha < 0}\\{\sin \beta > 0}\end{array}} \right.\).

Ta có \(\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3};\sin \beta = \sqrt {1 - {{\cos }^2}\beta } = \sqrt {1 - \frac{4}{9}} = \frac{{\sqrt 5 }}{3}\).

Suy ra \(\sin \left( {\alpha + \beta } \right) = \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) + \left( { - \frac{{2\sqrt 2 }}{3}} \right) \cdot \frac{{\sqrt 5 }}{3} = - \frac{{2 + 2\sqrt {10} }}{9}\).

Suy ra \(\sin \left( {\alpha + \beta } \right) = - \frac{{2 + 2\sqrt {10} }}{9} = - \frac{{2\left( {1 + \sqrt {10} } \right)}}{9}\).

Vậy\(a + b = 2 + 9 = 11\).

Đáp án:11.