1.已知二次三项式21x^2+ax-10可分解为两个整系数的一次因式之级,那么a是奇数?偶数?负数?

1.设21x^2+ax-10=(px+q)(mx+n) (p,q,m,n为整数)
原式=pmx^2+(pn+qm)x+qn
∴pm=21,pn+qm=a,qn=-10
∵21只能分解为两奇数乘积
∴p,m都为奇数
∵-10只能分解为1奇1偶乘积
∴q,n一奇一偶
∴pn+qm为奇数
∴a为奇数
a正负不确定
2.原式=(a-b)(a-c)=7
∵a＞b
∴a-b＞0
∵a.b.c是正整数
∴a-b=1,a-c=7或a-b=7,a-c=1
∴a-c=1或7
3.应该是(x+y）(x+y+2xy)+(xy+1)(xy-1)吧
设x+y=a,xy=b
原式=a(a+2b)+(b+1)(b-1)
=a^2+2ab+b^2-1
=(a+b)^-a
=(a+b-1)(a+b+1)
将x+y=a,xy=b代回
原式=(xy+x+y-1)(xy+x+y+1)
=(xy+x+y-1)(x+1)(y+1)
4.(x^2+y^2-2x+1)^2-(4y-4xy)(x^2-y^2-2x+1)
=(x^2+y^2-2x+1)^2-4y(1-x)(x^2-y^2-2x+1)
=[(x-1)^2+y^2]^2+4y(x-1)[(x-1)^2-y^2]
=[(x-1)^2+y^2]^2+4y(x-1)[(x-1)^2+y^2-2y^2]
=[(x-1)^2+y^2]^2+2*2y(x-1)[(x-1)^2+y^2]+4y^2(x-1)^2 -4y^2(x-1)^2-8y^3(x-1)
=[(x-1)^2+y^2+2y(x-1)]^2-4y^2(x-1)(x-1+2y)
=[(x-1)(x-1+2y)+y^2]^2-4y^2(x-1)(x-1+2y)
=[(x-1)(x-1+2y)]^2+2y^2(x-1)(x-1+2y)+y^4--4y^2(x-1)(x-1+2y)
=[(x-1)(x-1+2y)]^2-2y^2(x-1)(x-1+2y)+y^4
=[(x-1)(x-1+2y)+y^2]^2
=[(x-1)^2+2y(x-1)+y^2]^2
={[(x-1)+y]^2}^2
=(x+y-1)^4
5.x^3+ax^2+bx+c=K*(x^2+3x-4)
x=1代入得：a+b+c+1=0 ……（1）
x=-4代入得：16a-4b+c-64=0 ……（2）
4*(1)+(2),得：20a+5c-60=0,4a+c=12
(2)-6*(1),得：10a-10b-5c-70=0,2a-2b-c=14
4a+c=12,c大于等于a大于1,所以a=c=2,b=-5
6.∵a^2+ab-2b^2=0,
∴(a+2b)(a-b)=0,
∴a+2b=0或a-b=0,
∴a=-2b或a=b,
当a=-2b时,
(2a-b)/(2a+b)
=(-4b-b)/(-4b+b)
=5/3
当a=b时,
(2a-b)/(2a+b)
=b/(3b)
=1/3
∴原式=5/3或1/3