Question

计算：（2⁴+2²+1）×（4⁴+4²+1）×（6⁴+6²+1）×（8⁴+8²+1）×（10⁴+10²+1）÷（（3⁴+3²+1）×（5⁴+5²+1）×（7⁴+7²+1）×（9⁴+9²+1）×（11⁴+11²+1））．

Answer 1

考点：有理数无理数的概念与运算

专题：

分析：利用（x⁴+x²+1）=（x²+x+1）（x²-x+1）；x²-x+1=（x-1）²+（x-1）+1，进而将原式变形，进而化简求值即可．

解答：解：∵（x⁴+x²+1）=（x²+x+1）（x²-x+1）；x²-x+1=（x-1）²+（x-1）+1；
∴原式=

(22+2+1)(22-2+1)(42+4+1)(42-4+1)(62+6+1)(62-6+1)(82+8+1)(82-8+1)(102+10+1)(102-10+1)

(32+3+1)(32-3+1)(52+5+1)(52-5+1)(72+7+1)(72-7+1)(92+9+1)(92-9+1)(112+11+1)(112-11+1)

=

(22+2+1)(22-2+1)(42+4+1)(32+3+1)(62+6+1)(52+5+1)(82+8+1)(72+7+1)(102+10+1)(92+9+1)

(32+3+1)(22+2+1)(52+5+1)(42+4+1)(72+7+1)(62+6+1)(92+9+1)(82+8+1)(112+11+1)(102+10+1)

=

22-2+1

112+11+1

=

3

133

．

点评：此题主要考查了有理数无理数的概念与运算，正确化简各式是解题关键．