摘要:1.计算题: (1) , (2) (3) (4), (5) (6), (7) , (8)(p为整数).
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计算题:
(1)(-2a3b2c)3÷(4a2b3)2-a4c•(-2ac2);
(2)(3x2+4x-1)(3x2-4x-1);
(3)(a-2)(a2+a+1)+(a-
)(1-a)(a+
);
(4)(-2)4+(-
)-3+(-0.125)15×(215)3+22n÷4n-1.(n为正整数)
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(1)(-2a3b2c)3÷(4a2b3)2-a4c•(-2ac2);
(2)(3x2+4x-1)(3x2-4x-1);
(3)(a-2)(a2+a+1)+(a-
| 3 |
| 3 |
(4)(-2)4+(-
| 1 |
| 2 |