快,初二数学！！~~

1、（x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4)][(x+2)(x+3)]+1
=(x²+5x+4)(x²+5x+6)+1
=[(x²+5x)+4][(x²+5x)+6]+1
=(x²+5x)²+10(x²+5x)+24+1
=(x²+5x)²+10(x²+5x)+25
=[(x²+5x)+5]²
=(x²+5x+5)².
2、a²-2ab+2b²+4a+8=0
2a²-4ab+4b²+8a+16=0
(a²-4ab+4b²)+(a²+8a+16)=0
(a-2b)²+(a+4)²=0
a-2b=0且a+4=0
解得a=-4，b=-2
所以ab=-4×(-2)=8

3、x+a)(x+b)+c(x-10)=(x-11)(x+1)
x^2+(a+b+c)x+ab-10c=x^2-10x-11
(a+b+c+10)x=10c-ab-11
因为等式对任意x成立,所以
a+b+c+10=0
10c-ab-11=0
即
a+b=-10-c
ab=10c-11
a,b为方程x^2+(10+c)x+(10-11)=0的两整数解
所以▲=(10+c)^2-4(10c-11)为完全平方数
▲=100+20c+c^2-40c+44
=c^2-20c+144
=(c-10)^2+44
=K^2 (K为整数)

K^2-(c-10)^2=(K+c-10)(K-c+10)=44(1)
44=4*11=1*44=2*22(2)
K,c均为整数(3)
(K+c-10)+(K-c+10)=2K为偶数(4