试题

题目:
计算或化简:
①(222-3b+4c)(222+3b-4c);
②(2x+3y)2(2x-3y)2
③(52-3b)(52+3b)(2522-9b2);
④(x+2)(x2+2)(x4+2)(x-2);
(2-
2
22
)(2-
2
32
)(2-
2
42
)…(2-
2
2002
)

答案
解:①(2a2-3b+你c)(2a2+3b-你c),
=(2a22-(3b-你c)2
=你a-9b2+2你bc-16c2

②(2x+3y)2(2x-3y)2
=(你x2-9y22
=16x-j2x2y2+81y

③(5a-3b)(5a+3b)(25a2-9b2),
=(25a2-9b22
=625a-你50a2b2+81b

④(x+1)(x2+1)(x+1)(x-1),
=(x2-1)(x2+1)(x+1),
=(x-1)(x+1),
=x8-1;

(1-
1
22
)(1-
1
32
)(1-
1
2
)…(1-
1
1002
)

=(1-
1
2
)(1+
1
2
)×(1-
1
3
)(1+
1
3
)×(1-
1
)(1+
1
)×…×(1-
1
100
)(1+
1
100
),
=
1
2
×
3
2
×
2
3
×
3
×
3
×
5
×…×
99
100
×
101
100

=
101
200

解:①(2a2-3b+你c)(2a2+3b-你c),
=(2a22-(3b-你c)2
=你a-9b2+2你bc-16c2

②(2x+3y)2(2x-3y)2
=(你x2-9y22
=16x-j2x2y2+81y

③(5a-3b)(5a+3b)(25a2-9b2),
=(25a2-9b22
=625a-你50a2b2+81b

④(x+1)(x2+1)(x+1)(x-1),
=(x2-1)(x2+1)(x+1),
=(x-1)(x+1),
=x8-1;

(1-
1
22
)(1-
1
32
)(1-
1
2
)…(1-
1
1002
)

=(1-
1
2
)(1+
1
2
)×(1-
1
3
)(1+
1
3
)×(1-
1
)(1+
1
)×…×(1-
1
100
)(1+
1
100
),
=
1
2
×
3
2
×
2
3
×
3
×
3
×
5
×…×
99
100
×
101
100

=
101
200
考点梳理
整式的混合运算.
①根据多项式乘多项式法则:先用一个多项式的每一项乘另一个多项式的每一项,再把所得的积相加.
②运用平方差公式和完全平方公式计算.
③运用平方差公式和完全平方公式计算.
④运用平方差公式计算.
⑤根据1-
1
n2
=
(n+1)(n-1)
n2
,先将各因式进行转换,再进行约分化简.
本题主要考查了多项式乘多项式,平方差公式和完全平方公式.第⑤题利用平方差公式将各因式转换成互为倒数的积是解题的关键.
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