问题
解答题
计算: (1)
(2)(1
(3)-
(4)(-2)4-[-32-(1-23)×7]; (5)2x+3y-4(y-x); (6)a2b-[2(a2b-2ac2)-(2bc-4ac2)]. |
答案
(1)原式=
-1 2
-3 4 1 6
=
-6 12
-9 12 2 12
=-
;5 12
(2)原式=
×24-7 4
×24-1 3
×2413 6
=42-8-52
=-18;
(3)原式=-
×(-1 9
)×25 4
×(-8)27 5
=-30;
(4)原式=16-[-9-(-7)×7]
=16-[-9+49)
=16-40
=-24;
(5)原式=2x+3y-4y+4x
=6x-y;
(6)原式=a2b-[2a2b-4ac2-2bc+4ac2]
=a2b-2a2b+4ac2+2bc-4ac2
=-a2b+2bc.