初中八年级数学奥数关于因式分解

Question

请解答以下问题
1. 4(x+5)(x+6)(x+10)(x+12)-3x·x
2. xy(x·x-y·y)+yz(y·y-z·z)+zx(z·z-x·x)
3. 2x·x+3xy-2y·y-x+8y-6
最好有写过程

Answer 1

1.
4(x+5)(x+6)(x+10)(x+12)－3x^2
= 4*[(x+5)(x+12)][(x+6)(x+10)] - 3x^2
= 4*(x^2 + 60 + 17x)(x^2 + 60 + 16x) - 3x^2
= 4*[(x^2+60)^2 + 33x(x^2+60) + 272x^2)] - 3x^2
= 4(x^2+60)^2 + 132x(x^2+60) + 1085x^2
= [2(x^2 +60) + 35x][2(x^2 +60) + 31x]
= (2x^2 + 35x + 120)(2x^2 + 31x + 120)
= (2x^2 + 35x + 120)(2x + 15)(x+8)
2.
xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=x^3y-xy^3+y^3z-yz^3+z^3x-zx^3
=x^3(y-z)+y^3(z-x)+z^3(x-y)
因为x-y=(x-z)+(z-y),所以：
=x^3(y-z)+y^3(z-x)+z^3[(x-z)+(z-y)]
=(x^3-z^3)(y-z)+(y^3-z^3)(z-x)
=(x-z)(x^2+xz+z^2)(y-z)+(y-z)(y^2+yz+z^2)(z-x)
=(x-z)(y-z)(x^2+zx+z^2-y^2-yz-z^2)
=(x-z)(y-z)(x-y)(x+y+z)
3.
x^2+3xy-2y^2-x+8y-6
=(x+2y)(2x-y)-x+8y-6
=(x+2y)(2x-y)-(x-8y)-6
=(x+2y-2)(2x-y+3) （十字相乘法）
[也可用待定系数法令(x+2y)(2x-y)-(x-8y)-6=(x+2y+a)(2x-y+b) }

Answer 2

第三题：（x+2y-2)(2x-y+3)
2x^2+3xy-2y^2-x+8y-6=2x^2+4xy-4x-xy-2y^2+2y+3x+6y-6
=(x+2y-2)*2x-(x+2y-2)*y+(x+2y-2)*3=（x+2y-2)(2x-y+3)
第二题： (x-z)(y-z)(x-y)(x+y+z)
xy(x·x-y·y)+yz(y·y-z·z)+zx(z·z-x·x)
=xy(x^2-z^2+z^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=xy(x^2-z^2)+xy(z^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=(xy-zx)(x^2-z^2)+(xy-yz)(z^2-y^2)
=x(y-z)(x+z)(x-z)+y(x-z)(z-y)(z+y)
=(x-z)(y-z)(x-y)(x+y+z)