Question

Найти сумму восемь целых шесть двадцать первых+(-11,4)+две целых пять седьмых

Answer 1

$8\frac{6}{21}+(-11,4)+2\frac{5}{7}=-\frac{2}{5}=-0,4$

1) $8\frac{2}{7}+(-11,4)=8\frac{2}{7}-11,4=\frac{58}{7}-11\frac{4}{10}=\frac{58}{7}-11\frac{2}{5}=\frac{58}{7}-\frac{57}{5}=\frac{290-399}{35}=-\frac{109}{35}=-3\frac{4}{35}$

2) $-3\frac{4}{35}+2\frac{5}{7}=2\frac{5}{7}-3\frac{4}{35}=\frac{19}{7}-\frac{109}{35}=\frac{95-109}{35}=-\frac{14}{35}=-\frac{2}{5}=-0,4$

Answer 2

8 6/21 +(-11,4) +2 5/7= 

174/21 -114/10  +19/7=

174/21 -114/10  +57/21=

231/21 -114/10 =

2310/210-2394/210=

-84/210=-12/30=-4/6