作业帮已知a^2+2a-2012=0,求代数式[(a-2)/(a^2+2a)-(a-1)/(a^2+4a+4)]÷(a-4)/(a+2)的值
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 15:43:34
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已知a^2+2a-2012=0,求代数式[(a-2)/(a^2+2a)-(a-1)/(a^2+4a+4)]÷(a-4)/(a+2)的值
已知a^2+2a-2012=0,求代数式[(a-2)/(a^2+2a)-(a-1)/(a^2+4a+
4)]÷(a-4)/(a+2)的值
已知a^2+2a-2012=0,求代数式[(a-2)/(a^2+2a)-(a-1)/(a^2+4a+4)]÷(a-4)/(a+2)的值
[(a-2)/(a^2+2a)-(a-1)/(a^2+4a+4)]÷(a-4)/(a+2)
=[(a-2)/a(a+2)-(a-1)/(a+2)^2]x(a+2)/(a-4)
=[(a-2)(a+2)/a(a+2)^2-a(a-1)/a(a+2)^2]x(a+2)/(a-4)
=(a^2-4-a^2+a)/a(a+2)^2x(a+2)/(a-4)
=(a-4)/a(a+2)^2x(a+2)/(a-4)
=1/a(a+2)
=1/(a^2+2a)
=1/2012
值为：1/2012 
a²+2a-2012=0, a²+2a=2012
[(a-2)/(a²+2a)-(a-1)/(a²+4a+4)] / [(a-4)/(a+2)]
=[(a-2)/[a(a+2)]-(a-1)/(a+2)²] / [(a-4)/(a+2)]
={[(a-2)(a+2)-a(a-1)]/[a(a+2)²]} / [(a-4)/(a+2)]
={[a²-4-a²+a]/[a(a+2)²]}*(a+2)/(a-4)
=(a-4)/[a(a+2)]*1/(a-4)
=1/(a²+2a)
=1/2012
[(a-2)/(a²+2a)-(a-1)/(a²+4a+4)]÷(a-4)/(a+2)
=[(a-2)/(a²+2a)-(a-1)/(a+2)²]×(a+2)/(a-4)
=1/(a²+2a)
∵a²+2a-2012=0
即，a²+2a=2012
∴原式=1/2012