题目内容

计算题
(1)(2×102)×(3×103
(2)-15+(
1
2
)2+(π-3)0

(3)(2x)3-y3÷(16xy2
(4)(x-1)(x2+x+1)
(1)(2×102)×(3×103),
=2×3×102+3
=6×105

(2)-15+(
1
2
)2+(π-3)0

=-1+
1
4
+1,
=
1
4


(3)(2x)3•y3÷(16xy2),
=8x3•y3÷(16xy2),
=
1
2
x2y;

(4)(x-1)(x2+x+1),
=(x-1)(x2+x+1),
=x3+x2+x-x2-x-1,
=x3-1;
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