四道数学题急~~~

(1)....3a^2-9ab+a
=a(3a-9b+1)

(2)....(x+y)(x+y+2xy)+(xy+1)(xy-1)
=(x+y)^2+2xy(x+y)+x^2y^2-1
=(x+y+xy)^2-1
=(x+y+xy+1)(x+y+xy-1)
=[x(1+y)+(1+y)](x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)

(3)....1/2(a^2+b^2+c^2)-bc+ca-ab
=(1/2)(a^2+b^2+c^2-2bc+2ca-2ab)
=(1/2)[(a-b)^2+2c(a-b)+c^2]
=(1/2)(a-b+c)^2

解答:
三角形ABC的三边a.b.c有如下关系:
-c^2+a^2+2ab-2bc=0
(a-c)(a+c)+2b(a-c)=0
(a-c)(a+c+2b)=0
因为a.b.c是三角形ABC的三边
所以a+c+2b>0
所以a-c=0
a=c
所以这个三角形是等腰三角形