#6 с решением пожалуйста

Решите задачу:

$\frac{\left(13,75+9\frac16\right)\cdot1,2}{\left(10,3-8\frac12\right)\cdot\frac59}+\frac{\left(6,8-3\frac35\right)\cdot5\frac56}{\left(3\frac23-3\frac16\right)\cdot56}-27\frac16=\frac{\left(13\frac34+9\frac16\right)\cdot1\frac15}{\left(10\frac3{10}-8\frac12\right)\cdot\frac59}+\frac{\left(6\frac45-3\frac35\right)\cdot5\frac56}{\left(3\frac23-3\frac16\right)\cdot56}-27\frac16=$
$=\frac{\left(\frac{55}4+\frac{55}6\right)\cdot\frac65}{\left(\frac{103}{10}-\frac{17}2\right)\cdot\frac59}+\frac{\left(\frac{34}5-\frac{18}5\right)\cdot\frac{35}6}{\left(\frac{11}3-\frac{19}6\right)\cdot56}-\frac{163}6=\frac{\frac{165+110}{12}\cdot\frac65}{\frac{103-85}{10}\cdot\frac59}+\frac{\frac{34-18}5\cdot\frac{35}6}{\frac{22-19}3\cdot56}-\frac{163}6=$
$=\frac{\frac{275}{12}\cdot\frac65}{\frac{18}{10}\cdot\frac59}+\frac{\frac{16}5\cdot\frac{35}6}{\frac{3}6\cdot56}-\frac{163}6=\frac{\frac{55}{2}}{1}+\frac{\frac{56}3}{28}-\frac{223}6=\frac{55}2+\frac{56}{3\cdot28}-\frac{163}6=$
$=\frac{55}2+\frac23-\frac{163}6=\frac{165+4-163}{6}=\frac{6}{6}=1$