$$40-5 \times 2OF3+(19-3)$$
$56-10F_{3}O$
$$40-10OF_{3}+19-3$$
$$59-10OF_{3}-3$$
$$56-10OF_{3}$$
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$2\left(28-5F_{3}O\right)$
$$-10F_{3}O+56$$
$$2\left(-5F_{3}O+28\right)$$
