问题 解答题
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)
答案

(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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