题目内容
(1)已知10a=2,10b=5,10c=3,求103a-2b+c的值.
(2)计算:(2a
b
)2(-6a
b
)÷(-3a
b
)3.
(2)计算:(2a
| 2 |
| 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 2 |
| 3 |
| 1 |
| 6 |
| 5 |
| 6 |
分析:(1)(2)利用指数幂的运算性质即可得出;
解答:解:(1)103a-2b+c=
=
=
.
(2)原式=22(a
)2(b
)2(-6)a
b
÷[(-3)3(a
)3(b
)3
=
×a
+
-
b1+
-
=
a
b-
.
| (10a)3×10c |
| (10b)2 |
| 23×3 |
| 52 |
| 24 |
| 25 |
(2)原式=22(a
| 2 |
| 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 2 |
| 3 |
| 1 |
| 6 |
| 5 |
| 6 |
=
| 4×(-6) |
| -27 |
| 4 |
| 9 |
| 1 |
| 2 |
| 3 |
| 6 |
| 2 |
| 3 |
| 5 |
| 2 |
=
| 8 |
| 9 |
| 4 |
| 9 |
| 5 |
| 6 |
点评:熟练掌握指数幂的运算性质是解题的关键.
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