已知25a2-10a+1+|4b+1|=0,求[(4a+3b)(4a-3b)-(2a-5b)(8a+5b)]÷(-2b).

问题描述:

已知25a2-10a+1+|4b+1|=0,求[(4a+3b)(4a-3b)-(2a-5b)(8a+5b)]÷(-2b).

∵25a2-10a+1+|4b+1|=0,∴(5a-1)2+|4b+1|=0,
∴a=

1
5
、b=
1
4

∴[(4a+3b)(4a-3b)-(2a-5b)(8a+5b)]÷(-2b)
=(16a2-9b2-16a2+25b2+30ab)÷(-2b)
=(16b2+30ab)÷(-2b)
=-8b-15a
=-8×(-
1
4
)-15×
1
5

=2-3
=-1.
答案解析:由25a2-10a+1+|4b+1|=0可求得a、b的值,化简[(4a+3b)(4a-3b)-(2a-5b)(8a+5b)]÷(-2b),再代入计算即可.
考试点:整式的加减—化简求值;非负数的性质:绝对值;非负数的性质:偶次方.
知识点:本题利用方程25a2-10a+1+|4b+1|=0的特殊解法,解得a、b的值,从而使问题变的简单.