Question

两道初二数学化简题.化简：（1）a²+2a+1分之a²-4a+4·a²-4分之a+1（2）？

Answer 1

1、
(a²-4a+4)/(a²+2a+1)*(a+1)/(a²-4)
=(a-2) ²/(a+1) ²*(a+1)/(a+2)(a-2)
=(a-2)/(a+1)(a+2)
2、
(y-3)/(4y-8)*(y-2)/(y²-9)
=(y-3)/4(y-2)*(y-2)/(y+3)(y-3)
=1/4（y+3),5,a²+2a+1分之a²-4a+4·a²-4分之a+1
=(a-2)²/(a+1)²×(a+1)/(a+2)(a-2)
=(a-2)/(a+2)
4y-8分之y-3·y²-9分之y-2
=(y-3)/4(y-2)×(y-2)/(y+3)(y-3)
=1/(4y+12),0,(1)a²+2a+1分之a²-4a+4·a²-4分之a+1
=(a^2-4a+4/a^2+2a+1)*(a+1)/(a^2-4)
=(a-2)^2/(a+1)^2*(a+1)/(a+2)(a-2)
=(a-2)(a-2)/(a+1)(a+1)*(a+1)/(a+2)(a-2)
=(a-2)/(a+1)(a+2)
(2)
y-3/(4y-8)*y-2/(y^2-9)
(y-3)/4(y-2)*(y-2)/(y+3)(y-3)
=1/4(y+3),0,两道初二数学化简题.
化简：
（1）a²+2a+1分之a²-4a+4·a²-4分之a+1
（2）4y-8分之y-3·y²-9分之y-2