(1)以知:a+b=1ab=-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)

Question

(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)

Answer 1

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)]}+{(x-5)[(x-4)+(x-3)]}
=(x-2)(2x-7)+(x-5)(2x-7)
=(2x-7)[(x-2)+(x-5)]
=(2x-7)的平方
分也不多,随便帮帮你啦,不知道来得及吗?
楼上的第一题算错了,哪有你那样解2次方程的呀.