Решите пожалуйста: (21/113 - 14/19 + 7/8 - 28/41) + (4/41 - 1/8 + 2/19 - 3/113) : 1/7=?

Решите задачу:

$(\frac{21}{113}-\frac{14}{19}+\frac{7}{8}-\frac{28}{41})+(\frac{4}{41}-\frac{1}{8}+\frac{2}{19}-\frac{3}{113}):\frac{1}{7}$

$1) \frac{21}{113}-\frac{14}{19}= \frac{21*19-14*113}{113*19}= \frac{399-1582}{2147} = \frac{-1183}{2147}$

$2) \frac{7}{8}-\frac{28}{41}= \frac{7*41-28*8}{8*41} = \frac{287-224}{328} = \frac{63}{328}$

$3) \frac{-1183}{2147}+\frac{63}{328}= \frac{-1183*328+63*2147}{2147*328} = \frac{-388024+135261}{704216} = \frac{-252763}{704216}$

$4) \frac{4}{41}-\frac{1}{8}= \frac{4*8-1*41}{41*8} = \frac{32-41}{328}= \frac{-9}{328}$

$5) \frac{2}{19}-\frac{3}{113}= \frac{2*113-3*19}{19*113} = \frac{226-57}{2147} = \frac{169}{2147}$

$6) \frac{-9}{328}+\frac{169}{2147}= \frac{-9*2147+169*328}{328*2147} = \frac{-19323+55432}{704216} = \frac{36109}{704216}$

$7) \frac{36109}{704216}:\frac{1}{7}=\frac{36109}{704216}*7=\frac{252763}{704216}$

$8) \frac{-252763}{704216}+\frac{252763}{704216}=0$