题目内容
计算题(1)(2×102)×(3×103)
(2)-15+(
| 1 | 2 |
(3)(2x)3•y3÷(16xy2)
(4)(x-1)(x2+x+1)
分析:(1)根据单项式乘单项式的法则计算;
(2)先算乘方,再算加减;
(3)先算幂的乘方,再根据单项式的乘法,单项式的除法法则计算;
(4)根据多项式乘法,先用单项式乘以多项式的每一项,再把所得的积相加计算即可.
(2)先算乘方,再算加减;
(3)先算幂的乘方,再根据单项式的乘法,单项式的除法法则计算;
(4)根据多项式乘法,先用单项式乘以多项式的每一项,再把所得的积相加计算即可.
解答:解:(1)(2×102)×(3×103),
=2×3×102+3,
=6×105;
(2)-15+(
)2+(π-3)0,
=-1+
+1,
=
;
(3)(2x)3•y3÷(16xy2),
=8x3•y3÷(16xy2),
=
x2y;
(4)(x-1)(x2+x+1),
=(x-1)(x2+x+1),
=x3+x2+x-x2-x-1,
=x3-1;
=2×3×102+3,
=6×105;
(2)-15+(
| 1 |
| 2 |
=-1+
| 1 |
| 4 |
=
| 1 |
| 4 |
(3)(2x)3•y3÷(16xy2),
=8x3•y3÷(16xy2),
=
| 1 |
| 2 |
(4)(x-1)(x2+x+1),
=(x-1)(x2+x+1),
=x3+x2+x-x2-x-1,
=x3-1;
点评:此题考查整式的混合运算,主要考查平方、立方及整式的乘除运算法则,熟练掌握运算法则是解题的关键.
练习册系列答案
相关题目