搜搜考试网求助：几道因式分解题⑴(x+y)^4+(x²-y²）²+(x-y)^4⑵（x^4-4x&s
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求助：几道因式分解题
⑴(x+y)^4+(x²-y²）²+(x-y)^4
⑵（x^4-4x²+1）（x^4+3x²+1）+10x^4
⑶16（6x-1）（2x-1）（3x+1)（x-1)+25
⑷a³b-ab³+a²+b²+1
⑸（x²-y²）²-8（x²+y²）+16
⑹（a+b-2ab）（a+b-2）+（1-ab）²
指数不好打,^4代表4次方.
⑴(x+y)^4+(x²-y²）²+(x-y)^4
⑵（x^4-4x²+1）（x^4+3x²+1）+10x^4
⑶16（6x-1）（2x-1）（3x+1)（x-1)+25
⑷a³b-ab³+a²+b²+1
⑸（x²-y²）²-8（x²+y²）+16
⑹（a+b-2ab）（a+b-2）+（1-ab）²
指数不好打,^4代表4次方.
(1)原式=(x+y)^4+(x+y)^2(x-y)^2+(x-y)^4
=(x+y)^4+2(x+y)^2(x-y)^2+(x-y)^4-(x+y)^2(x-y)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=(2x^2+2y^2)^2-(x^2-y^2)^2
=[(2x^2+2y^2)+(x^2-y^2)][(2x^2+2y^2)-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)
(2)为方便起见,可以用换元法来分解,设x^4+1=M,x^2=N,
则原式 =(M-4N)(M+3N)+10N^2
=M^2-MN-12N^2+10N^2
=M^2-MN-2N^2
=(M-2N)(M+N)
=(x^4+1-2x^2)(x^4+1+x^2)
=(x^2-1)^2×[(x^4+2x^2+1)-x^2]
=(x-1)^2×(x+1)^2×[(x^2+1)^2-x^2 ]
=(x-1)^2×(x+1)^2×(x^2+1-x)×(x^2+1+x)
(3)原式=(6x-1)(4x-2)(6x+2)(4x-4)+25
=(24x^2-16x+2) (24x^2-16x-8)+25
设24x^2-16x+2=t,
则原式=t(t-10)+25=(t-5)^2=(24x^2-16x-3)^2
(4)原式=a^3b-ab^3+a^2+b^2+1+ab-ab
=(a^3b-ab^3)+(a^2-ab)+(ab+b^2+1)
=ab(a+b)(a-b)+a(a-b)+(ab+b^2+1)
=a(a-b)〔b(a+b)+1]+(ab+b2+1)
=[a(a-b)+1](ab+b2+1)
=(a2-ab+1)(b2+ab+1)．
(5)原式=(x+y)^2(x-y)^2-8(x^2+y^2)+16
=((x+y)^2-4)((x-y)^2-4)
=(x+y+2)*(x+y-2)*(x-y+2)*(x-y-2)
(6)原式=[(a+b)-2ab][(a+b)-2]+(1-ab)(1-ab)
=(a+b)^2-2(ab+1)(a+b)+4ab+(1-ab)^2
=(a+b)^2-2(ab+1)(a+b)+[4ab+(1-ab)^2]
=(a+b)^2-2(ab+1)(a+b)+(1+ab)^2
=[(a+b)-(ab+1)]^2
=[(a-1)(1-b)]^2
=(a-1)^2*(b-1)^2
=(x+y)^4+2(x+y)^2(x-y)^2+(x-y)^4-(x+y)^2(x-y)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=(2x^2+2y^2)^2-(x^2-y^2)^2
=[(2x^2+2y^2)+(x^2-y^2)][(2x^2+2y^2)-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)
(2)为方便起见,可以用换元法来分解,设x^4+1=M,x^2=N,
则原式 =(M-4N)(M+3N)+10N^2
=M^2-MN-12N^2+10N^2
=M^2-MN-2N^2
=(M-2N)(M+N)
=(x^4+1-2x^2)(x^4+1+x^2)
=(x^2-1)^2×[(x^4+2x^2+1)-x^2]
=(x-1)^2×(x+1)^2×[(x^2+1)^2-x^2 ]
=(x-1)^2×(x+1)^2×(x^2+1-x)×(x^2+1+x)
(3)原式=(6x-1)(4x-2)(6x+2)(4x-4)+25
=(24x^2-16x+2) (24x^2-16x-8)+25
设24x^2-16x+2=t,
则原式=t(t-10)+25=(t-5)^2=(24x^2-16x-3)^2
(4)原式=a^3b-ab^3+a^2+b^2+1+ab-ab
=(a^3b-ab^3)+(a^2-ab)+(ab+b^2+1)
=ab(a+b)(a-b)+a(a-b)+(ab+b^2+1)
=a(a-b)〔b(a+b)+1]+(ab+b2+1)
=[a(a-b)+1](ab+b2+1)
=(a2-ab+1)(b2+ab+1)．
(5)原式=(x+y)^2(x-y)^2-8(x^2+y^2)+16
=((x+y)^2-4)((x-y)^2-4)
=(x+y+2)*(x+y-2)*(x-y+2)*(x-y-2)
(6)原式=[(a+b)-2ab][(a+b)-2]+(1-ab)(1-ab)
=(a+b)^2-2(ab+1)(a+b)+4ab+(1-ab)^2
=(a+b)^2-2(ab+1)(a+b)+[4ab+(1-ab)^2]
=(a+b)^2-2(ab+1)(a+b)+(1+ab)^2
=[(a+b)-(ab+1)]^2
=[(a-1)(1-b)]^2
=(a-1)^2*(b-1)^2
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