Помогите пожалуйста решить примеры

А)
$(800-981 * \frac{5}{8} ) * \frac{1}{5} = 800 * \frac{1}{5} - 981 * \frac{5}{8} * \frac{1}{5} = \\ \\ = \frac{800*1}{1*5} - \frac{981*5*1}{1*8*5} = \frac{800}{5} - \frac{981}{8} = \\ \\ =160- 122 \frac{5}{8} =37 \frac{3}{8}$

$800-981 * \frac{5}{8} * \frac{1}{5} = 800- \frac{981}{8} = 800-122 \frac{5}{8} =677 \frac{3}{8}$

б)
$4* \frac{1}{4}+ \frac{1}{8} -1 = \frac{4}{4} + \frac{1}{8} -1= 1 \frac{1}{8} -1=\frac{1}{8}$

$4*( \frac{1}{4}+ \frac{1}{8}) -1= 4*( \frac{2+1}{8})-1= \\ \\ = \frac{4*3}{1*8} -1= \frac{3}{2} - \frac{2}{2} = \frac{1}{2}$

в)
$3:6+ \frac{1}{4} *(3- \frac{1}{2} )= \frac{1}{2} + \frac{1}{4} * 2 \frac{1}{2} = \\ \\ = \frac{1}{2} + \frac{1*5}{4*2} = \frac{4+5}{8} = 1\frac{1}{8}$

$3:6+ \frac{1}{4} *3- \frac{1}{2} = \frac{1}{2} + \frac{3}{4} - \frac{1}{2} = \frac{3}{4}$

г)
$2:12 +( \frac{5}{6} - \frac{7}{12} ) * \frac{1}{7} = \frac{2}{12} +( \frac{10-7}{12} )* \frac{1}{7} = \\ \\ =\frac{1}{6} + \frac{3*1}{12*7} = \frac{1}{6} + \frac{1*1}{6*7} = \\ \\ = \frac{7}{42} + \frac{1}{42} = \frac{8}{42} = \frac{4}{21}$

$2:12 + \frac{5}{6} - \frac{7}{12} * \frac{1}{7}= \frac{1}{6} + \frac{5}{6} - \frac{1}{12} = 1- \frac{1}{12} = \frac{11}{12}$