Question

八年级下册数学先化简再求值问题

题目
（a-2/a²-2a - a-1/a²+4a+4）÷ a-4/a+2
其中a满足a²+2a+1=0
怎么解，详细的解题步骤

Answer 1

a²+2a-1=0
a²+2a=1

原式=[(a-2)/a(a+2)-(a-1)/(a+2)²]×(a+2)/(a-4)
={[(a-2)(a+2)-a(a-1)]/a(a+2)²}×(a+2)/(a-4)
=(a²-4-a²+a)(a+2)/[a(a-4)(a+2)²]
=(a-24)(a+2)/[a(a-4)(a+2)²]
=1/(a²+2a)
=1/1
=1

Answer 2

a²+2a+1=（a+1）^2=0
a=-1
代入上式
（a-2/a²-2a - a-1/a²+4a+4）÷ a-4/a+2
=（-3+2+2-4+4）÷ (-5)/1
=1/(-5)
=-1/5

Answer 3

数学就是要多练习哦，这是中考的必考类型题目，还有就是对分解因式、通风、约分要很熟悉哦。希望你加油哦