Ta có:
$A = \frac{\frac{-1}{2} -5 \times \frac{3}{2}^{2}}{15\frac{2}{9} + (-\frac{2}{3})^{2}}$
= $A = \frac{\frac{-1}{2} -5 \times \frac{9}{4}}{\frac{137}{9} + \frac{4}{9}} = \frac{\frac{-1}{2} - \frac{45}{4}}{\frac{47}{3}} = \frac{\frac{-47}{4}}{\frac{47}{3}}$
$= \frac{-47}{4} / \frac{47}{3} = \frac{-3}{4}.$
$B = \frac{7}{12}\times 3.4 - \frac{7}{12}\times 8.8.$
$= \frac{7}{12} \times (3.4 - 8.8) = \frac{7}{12} \times (-5.4) = \frac{7}{12} \times \frac{-27}{5} = \frac{-63}{20}.$
Do đó $A - 5B = \frac{-3}{4} -5 \times \frac{-63}{20} = \frac{-3}{4} + \frac{63}{4} = 15$