初二数学：两道因式分解题!把下列各式因式分解：（1） （a^2+a+1）(a^2-6a+1)+12a^2 (2) (x+y)^4+(x^2-y^2)^2+(x-y)^4重点要详细的过程!
2019-04-18
初二数学：两道因式分解题!
把下列各式因式分解：
（1） （a^2+a+1）(a^2-6a+1)+12a^2
(2) (x+y)^4+(x^2-y^2)^2+(x-y)^4
重点要详细的过程!
优质解答
1、(a^2+a+1)(a^2-6a+1)+12a^2
=[(a^2+1)+a][(a^2+1)-6a]+12a^2
=(a^2+1)^2-5a*(a^2+1)+6a^2
=(a^2-3a+1)(a^2-2a+1)
=[a-(3+根号5)/2][a-(3-根号5)/2](a-1)^2；
2、(x+y)^4+(x^2-y^2)^2+(x-y)^4
=[(x+y)^4+2(x^2-y^2)^2+(x-y)^4]-(x^2-y^2)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=[(x+y)^2+(x-y)^2+(x^2-y^2)][(x+y)^2+(x-y)^2-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)
1、(a^2+a+1)(a^2-6a+1)+12a^2
=[(a^2+1)+a][(a^2+1)-6a]+12a^2
=(a^2+1)^2-5a*(a^2+1)+6a^2
=(a^2-3a+1)(a^2-2a+1)
=[a-(3+根号5)/2][a-(3-根号5)/2](a-1)^2；
2、(x+y)^4+(x^2-y^2)^2+(x-y)^4
=[(x+y)^4+2(x^2-y^2)^2+(x-y)^4]-(x^2-y^2)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=[(x+y)^2+(x-y)^2+(x^2-y^2)][(x+y)^2+(x-y)^2-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)