题目内容
计算:(1)3(2m-| 1 | 3 |
(2)-7y+(2y-3)-2(3y+2).
分析:根据有理数乘法法则、整式的加减法则计算.
解答:解:(1)3(2m-
)
=3×2m-3×
=6m-1.
(2)-7y+(2y-3)-2(3y+2)
=-7y+2y-3-2×3y+(-2)×2
=-7y+2y-3-6y-4
=(-7+2-6)y-7
=-11y-7.
| 1 |
| 3 |
=3×2m-3×
| 1 |
| 3 |
=6m-1.
(2)-7y+(2y-3)-2(3y+2)
=-7y+2y-3-2×3y+(-2)×2
=-7y+2y-3-6y-4
=(-7+2-6)y-7
=-11y-7.
点评:不为零的有理数相乘的法则:两数相乘,同号得正,异号得负,并把绝对值相乘.
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