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:2024/07/15 06:48:03
![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](/uploads/image/z/5282587-19-7.jpg?t=%E5%B7%B2%E7%9F%A5a%5E2%2B2a-2012%3D0%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F%5B%28a-2%29%2F%28a%5E2%2B2a%29-%28a-1%29%2F%28a%5E2%2B4a%2B4%29%5D%C3%B7%28a-4%29%2F%28a%2B2%29%E7%9A%84%E5%80%BC)
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