樓主太強(qiáng)大了,..這是初一嗎?我盡量把這當(dāng)做一次挑戰(zhàn)來(lái)檢驗(yàn)自己的學(xué)成.
1.原式=(3x^5+3x^4)-(6x^4+6x^3)-(7x^3+7x^2)-(4x^2+4x)-(6x+6)
=3x^4(x+1)-6x^3(x+1)-7x^2(x+1)-4x(x+1)-6(x+1)
=(3x^4-6x^3-7x^2-4x-6)(x+1)
=[3x^4+3x^3-9x^3-9x^2+2x^2+2x-6x-6](x+1)
=[3x^3(x+1)-9x^2(x+1)+2x(x+1)-6(x+1)](x+1)
=(3x^3-9x^2+2x-6)(x+1)^2
=[3x^2(x-3)+2(x-3)](x+1)^2
=(3x^2+2)(x-3)(x+1)^2
2.取x+2=A,則
原式=(A-1)^4+(A+1)^4-272
=2A^4+12A^2-270
=2(A^4+6A^2-135)
=2(A^2+15)(A^2-9)
=2(x^2+4x+19)(A-3)(A+3)
=2(x^2+4x+19)(x-1)(x+5)
3.取ab=A,a+b=B
則原式=(A-1)^2+(B-2)(B-2A)
=A^2-2A+1+B^2-2AB-2B+4A
=A^2-2(B-1)A+B^2-2B+1
=A^2-2(B-1)A+(B-1)^2
=(A-B+1)^2
=(ab-a-b+1)^2
=(a-1)^2(b-1)^2
4.原式=a^2b^2-a^2c^2-c^2(ab+b^2-ac-bc)
=a^2b^2-a^2c^2-abc^2-b^2c^2+ac^3+bc^3
=(a^2-c^2)b^2+(c^3-ac^2)b+ac^2(c-a)
=(a+c)(a-c)b^2+c^2(c-a)b+ac^2(c-a)
=(a-c)[(a+c)b^2-c^2b-ac^2]
=(a-c)[(a+c)b+ac)(b-c)
=(a-c)(b-c)(ab+bc+ac)
5.原式=x^3+x^2+(5x^2+11x+6)
=x^2(x+1)+(5x+6)(x+1)
=(x^2+5x+6)(x+1)
=(x+1)(x+2)(x+3)
6.原式=(x^2+x)^2+2(x^2+x)+1
=(x^2+x+1)^2
7.原式=-[x^4+2x^2y+3y^2-8y+4]
=-[3y^2+(2x^2-8)y+(2-x^2)(2+x^2)]
=-(y+x^2+2)(3y-x^2+2)
=(x^2+y+2)(x^2-3y-2)
8.不好意思,樓主你能把這題寫(xiě)清楚點(diǎn)嗎?
9.原式=x^5-x^4-(2x^4-2x^3)-(3x^3-3x^2)+(8x^2-8x)-(4x-4)
=x^4(x-1)-2x^3(x-1)-3x^2(x-1)+8x(x-1)-4(x-1)
=(x^4-2x^3-3x^2+8x-4)(x-1)
=[x^4-x^3-(x^3-x^2)-(4x^2-4x)+4x-4](x-1)
=[x^3(x-1)-x^2(x-1)-4x(x-1)+4(x-1)](x-1)
=(x^3-x^2-4x+4)(x-1)^2
=[x^2(x-1)-4(x-1)](x-1)^2
=(x-1)^3(x^2-4)
=(x+2)(x-2)(x-1)^3
10.原式=(x^2+x-3)^2-2(x^2+x-3)-3
=(x^2+x-3-3)(x^2+x-3+1)
=(x^2+x-6)(x^2+x-2)
=(x-2)(x+3)(x-1)(x+2)
11.原式=2x^2(x-5)+3(x-5)
=(x-5)(2x^2+3)
.終于答完了..好累.不過(guò)希望樓主這么強(qiáng)大的人不要驕傲,這些題在一些情況下并不是難題,碰多元分解便用主元法,高次分解便用式根法,因式分解是有規(guī)律的