(1)以知:a+b=1 ab=-0.5
问题描述：
(1)以知:a+b=1 ab=-0.5
就a(a+b)*(a-b)-a*(a+b)的2次方
(2) 用简便方程计算.
(x-2)*(x-3)+(x-4)*(x-5)+(x-2)*(x-4)+(x-3)*(x-5)
答
一a（a+b)(a-b)-a(a+b)(a+b)=a(a+b)[(a-b)-(a+b)]=a(a+b)[a-b-a+-b]=-2ab(a+b)=-2*0.5*1=-1二(x-2)(x-3)+(x-4)(x-5)+(x-2)(x-4)+(x-3)(x-5)=[(x-2)(x-3)+)+(x-2)(x-4)]+[(x-4)(x-5)+(x-3)(x-5)]={(x-2)[(x-3)+(x-4)]...