问题 解答题
计算:
(1)(2-3)0-(
1
2
)-2+(
1
4
)2011×(-4)2011

(2)2(a43+(-2a32-(-a23+a2a10
(3)(-2x2y)3+(x22-[(-3x)2-y3-x3]
(4)(2x-1)(2x+1)(4x2-1)
答案

(1)原式=1-4-(

1
4
×4)2011

=1-4-1

=-4;

(2)原式=2a12+(4a6)•(-a6)+a12

=2a12-4a12+a12

=-a12

(3)原式=-8x6y3+x4•[9x2•y3-x3]

=x6y3-x7

(4)原式=(4x2-1)2

=16x4-8x2+1.

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