(1)(
| 3 |
| 2 |
| 1 |
| 2 |
(2)(m-3n)-2•(2m-2n-3)-2
(3)a-2b2•(-2a2b-2)-2÷(a-4b2)
(4)(2m2n-3)3(-mn-2)-2. 零指数幂(负指数幂和指数为1)
答案:
(1)(| 3 |
| 2 |
| 1 |
| 2 |
=1-4
=-3;(2)(m-3n)-2•(2m-2n-3)-2
=m6n-2•2-2m4n6
=
| 1 |
| 4 |
=
| 1 |
| 4 |
=a-2b2•(-2)-2a-4b4÷(a-4b2)
=
| 1 |
| 4 |
=
| 1 |
| 4 |
=
| b4 |
| 4a2 |
=23m6n-9(-m)-2n4
=8m6-2n-9+4
=8m4n-5
=
| 8m4 |
| n5 |



