题目内容
计算:
(1)-1-1÷32×
+2;
(2)(-3)×(-2)2-(-1)99÷
;
(3)(-10)2-5×(-3×2)2+23×10.
(1)-1-1÷32×
| 1 |
| 32 |
(2)(-3)×(-2)2-(-1)99÷
| 1 |
| 2 |
(3)(-10)2-5×(-3×2)2+23×10.
(1)-1-1÷32×
+2
=-1-1×
×
+2
=-1-
+2
=-
=
;
(2)(-3)×(-2)2-(-1)99÷
=(-3)×4-(-1)×2
=-12-(-2)
=-12+2
=-10;
(3)(-10)2-5×(-3×2)2+23×10
=100-5×(-6)2+8×10
=100-5×36+80
=100-180+80
=0.
| 1 |
| 32 |
=-1-1×
| 1 |
| 9 |
| 1 |
| 9 |
=-1-
| 1 |
| 81 |
=-
| 81+1-162 |
| 81 |
=
| 80 |
| 81 |
(2)(-3)×(-2)2-(-1)99÷
| 1 |
| 2 |
=(-3)×4-(-1)×2
=-12-(-2)
=-12+2
=-10;
(3)(-10)2-5×(-3×2)2+23×10
=100-5×(-6)2+8×10
=100-5×36+80
=100-180+80
=0.
练习册系列答案
相关题目