计算：(a+2a2−2a−a−1a2−4a+4)÷4−aa2−2a．

Question

计算：((a+2)/(a2−2a)−(a−1)/(a2−4a+4))÷(4−a)/(a2−2a)．

Answer 1

原式=[(a+2)/(a(a−2))-(a−1)/((a−2)2)]÷(4−a)/(a(a−2))
=[(a+2)/(a(a−2))-(a−1)/((a−2)2)]•(a(a−2))/(4−a)
=(a+2)/(4−a)-(a(a−1))/(a−2)×(1)/(4−a)
=((a+2)(a−2)−a(a−1))/((4−a)(a−2))
=(a−4)/((4−a)(a−2))
=-(a−4)/((a−4)(a−2))
=-(1)/(a−2)．