Question

Calculati:
a) [100-5*(473-4211)]*2+3*(273-21*12=
b)[36+7*(931*256*3)]*(18*2+44)-329=
c)3*{41+5[(46-21*2)*(7*9+37)-183]}+403*12=
d)18*{60+2*{9*(329-8*40)-37]}+12356-803*=

Answer 1

$a).[100-5\cdot(473-4211)]\cdot2+3\cdot(273-21\cdot12)= \\ =[100-5 \cdot (-3738)] \cdot 2+3 \cdot (273-252)= \\ =(100+18690) \cdot 2+3 \cdot 21= \\ =18790 \cdot 2+3 \cdot 21= \\ =37580+63= \\ =37643$
$b).[36+7\cdot(931\cdot256\cdot3)]\cdot(18\cdot2+44)-329= \\ =[36+7 \cdot (238336\cdot 3)] \cdot (36 +44)-329= \\ =(36+7 \cdot 715008) \cdot 80-329= \\ =(36+5005056) \cdot 80-329= \\ =5005092 \cdot 80-329= \\ =400407360-329= \\ =400407031$
$c).3 \cdot \{41+5[(46-21 \cdot 2)\cdot(7\cdot9+37)-183]\}+403 \cdot 12= \\ =3 \cdot \{41+5[(46-42) \cdot (63+37)-183]\}+403 \cdot 12= \\ =3 \cdot [41+5(4 \cdot 100-183)]+403 \cdot 12= \\ =3 \cdot [41+5(400-183)]+403 \cdot 12= \\ =3 \cdot (41+5 \cdot 217)+403 \cdot 12= \\ =3 \cdot (41+1085)+4836= \\ =3 \cdot 1126+4836= \\ =3378+4836= \\ =8214$
$d).18 \cdot \{60+2\cdot[9 \cdot (329-8 \cdot 40)-37]\}+12356-803= \\ =18\cdot\{60+2 \cdot [9\cdot (329-320)-37]\}+12356-803= \\ =18\cdot [60+2 \cdot (9 \cdot 9-37)+12356-803= \\ =18\cdot [60+2 \cdot (81-37)+12356-803= \\ =18 \cdot (60+2 \cdot 44)+12356-803= \\ =18 \cdot (60+88)+12356-803= \\ =18 \cdot 148+12356-803= \\ =2664+12356-803= \\ =15020-803= \\ =14217$