试题

题目:
(1)-22×7-(-3)×6+8
(2)(-1)2009-(
3
4
-
1
6
-
3
8
)×24-(-2)2×3
(3)5abc-2a2b-[3abc-3(4ab2+a2b)]
(4)
1
4
(-4x2+2x-8)-2(
1
2
x-1)
(5)(2x3-xyz)-2(x3-y3+xyz)+(xyz-2y3
(6)(2x2+x)-2[x2-2(3x2-x)].
答案
解:(1)原式=-4×7+3×6+8,
=-28+18+8,
=-2;

(2)原式=-1-
3
4
×24+
1
6
×24+
3
8
×24-4×3,
=-1-18+4+9-12,
=-18;

(3)原式=5abc-2a2b-[3abc-12ab2-3a2b],
=5abc-2a2b-3abc+12ab2+3a2b,
=2abc+a2b+12ab2

(4)原式=-x2+
1
2
x-2-x+2,
=-x2-
1
2
x;

(5)原式=2x3-xyz-2x3+2y3-2xyz+xyz-2y3=-2xyz;

(6)原式=2x2+x-2[x2-6x2+2x],
=2x2+x-2x2+12x2-4x,
=12x2-3x.
解:(1)原式=-4×7+3×6+8,
=-28+18+8,
=-2;

(2)原式=-1-
3
4
×24+
1
6
×24+
3
8
×24-4×3,
=-1-18+4+9-12,
=-18;

(3)原式=5abc-2a2b-[3abc-12ab2-3a2b],
=5abc-2a2b-3abc+12ab2+3a2b,
=2abc+a2b+12ab2

(4)原式=-x2+
1
2
x-2-x+2,
=-x2-
1
2
x;

(5)原式=2x3-xyz-2x3+2y3-2xyz+xyz-2y3=-2xyz;

(6)原式=2x2+x-2[x2-6x2+2x],
=2x2+x-2x2+12x2-4x,
=12x2-3x.
考点梳理
整式的加减;有理数的混合运算.
(1)先进行幂的运算,然后根据先乘除后加减的法则运算即可.
(2)先进行幂的运算,然后运算乘法分配律计算,然后再进行有理数的加减运算.
(3)先去小括号,再去中括号,然后合并同类项即可.
(4)先去括号,然后合并同类项即可.
(5)先去括号,然后合并同类项即可.
(6)先去小括号,再去中括号,然后合并同类项即可.
本题考查了有理数的混合运算及整式的加减,难度不大,关键是要在熟练运算步骤的基础上加强细心程度.
计算题.
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