數學知識交流---求值

(a-b) (c-d) / ( (b-c) (d-a) ) = 3因 (a-b) + (c-d) + (b-c) + (d-a) = 0令 (a-b) = x , (c-d) = y , (b-c) = z , 則 (d-a) = - (x+y+z)==>x y / - [z(x+y+z)] = 3故(a-c) (b-d) / ( (a-b)(c-d) )= (x+z) (y+z) / (xy)= (xy + zy + xz + z²) / (xy)= ( xy + z(x+y+z) ) / (xy)= 1 + z(x+y+z) / (xy)= 1 - 1/3= 2/3

[(a-b)(c-d)]/[(b-c)(d-a)]=3
=>[(b-c)(d-a)]/[(a-b)(c-d)]=1/3 --(1)
Therefore,
[(a-c)(b-d)] / [(a-b)(c-d)]
= [ab+cd-bc-ad] / [(a-b)(c-d)]
= [(ab+cd-bd-ac)+(bd+ac-bc-ad)] / [(a-b)(c-d)]
= [-(b-c)(d-a)+(c-d)(a-b)] / [(a-b)(c-d)]
= -1/3+1 (by (1))
= 2/3