数学问题:分解因式(有详细过程) 。。。。。。急急急!
1.-15x^4-5x^6+20x^22.a^7+a^2b^5-a^5b^2-b^73.7x^2-3y+xy-21x4.x^3+3x^2+3x+95.m^2+2mn-3m...
1. -15x^4-5x^6+20 x^2
2. a^7+a^2b^5-a^5b^2-b^7
3. 7x^2-3y+xy-21x
4. x^3+3x^2+3x+9
5. m^2+2mn-3m-3n+n^2
6. m^2 –n^2–2m+1
7. ab(c^2+d^2)+cd(a^2+b^2)
8. x^2-y^2+4x+2y+3
9. x^4+4
10. a^2 (b-c)+b^2 (c-a)+c^2 (a-b) 展开
2. a^7+a^2b^5-a^5b^2-b^7
3. 7x^2-3y+xy-21x
4. x^3+3x^2+3x+9
5. m^2+2mn-3m-3n+n^2
6. m^2 –n^2–2m+1
7. ab(c^2+d^2)+cd(a^2+b^2)
8. x^2-y^2+4x+2y+3
9. x^4+4
10. a^2 (b-c)+b^2 (c-a)+c^2 (a-b) 展开
展开全部
1. 解:原式 = -5x^6-15x^4+20 x^2
= -5 x^2(x^4+3x^2-4)
= -5 x^2(x^2+4)(x^2-1)
= -5 x^2(x^2+4)(x+1)(x-1)
2. 解:原式 = (a^7-a^5b^2)+(a^2b^5-b^7)
= a^5(a^2-b^2)+b^5(a^2-b^2)
= (a^2-b^2)(a^5+b^5)
= (a+b)(a-b)(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)
= (a+b)^2(a-b)(a^4-a^3b+a^2b^2-ab^3+b^4)
3. 解:原式=(7x^2-21x)+(xy-3y)
=7x(x-3)+y(x-3)
=(x-3)(7x+y)
4. 解:原式=(x^3+3x^2)+(3x+9)
=x^2(x+3)+3(x+3)
=(x+3)(x^2+3)
5. 解:原式=(m^2+2mn+n^2)+(-3m-3n)
=(m+n)^2-3(m+n)
=(m+n)(m+n-3)
6. 解:原式=(m^2–2m+1) –n^2
= (m–1)^2–n^2
= (m + n–1)( m–n–1)
7. 解:原式=abc^2+abd^2+cda^2+cdb^2
=(abc^2+cda^2)+(cdb^2+abd^2)
=ac(bc+ad)+bd(bc+ad)
=(bc+ad)(ac+bd)
8. 解:原式=x^2-y^2+4x+2y+4-1
=(x^2+4x+4)+(-y^2+2y-1)
=(x+2)^2-(y-1)^2
=(x+y+1)(x-y+3)
9. 解:原式=x^4+4x^2-4x^4+4
=(x^2+2)^2-(2x)^2
=(x^2+2x+2)(x^2-2x+2)
10. 解:原式= a^2(b-c)+b^2c-ab^2+ac^2-bc^2
= a^2(b-c)-a(b^2-c^2)+(b^2c-bc^2)
= (b-c) [a^2 -a(b+c)+bc]
= (b-c)(a-b)(a-c)
= -5 x^2(x^4+3x^2-4)
= -5 x^2(x^2+4)(x^2-1)
= -5 x^2(x^2+4)(x+1)(x-1)
2. 解:原式 = (a^7-a^5b^2)+(a^2b^5-b^7)
= a^5(a^2-b^2)+b^5(a^2-b^2)
= (a^2-b^2)(a^5+b^5)
= (a+b)(a-b)(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)
= (a+b)^2(a-b)(a^4-a^3b+a^2b^2-ab^3+b^4)
3. 解:原式=(7x^2-21x)+(xy-3y)
=7x(x-3)+y(x-3)
=(x-3)(7x+y)
4. 解:原式=(x^3+3x^2)+(3x+9)
=x^2(x+3)+3(x+3)
=(x+3)(x^2+3)
5. 解:原式=(m^2+2mn+n^2)+(-3m-3n)
=(m+n)^2-3(m+n)
=(m+n)(m+n-3)
6. 解:原式=(m^2–2m+1) –n^2
= (m–1)^2–n^2
= (m + n–1)( m–n–1)
7. 解:原式=abc^2+abd^2+cda^2+cdb^2
=(abc^2+cda^2)+(cdb^2+abd^2)
=ac(bc+ad)+bd(bc+ad)
=(bc+ad)(ac+bd)
8. 解:原式=x^2-y^2+4x+2y+4-1
=(x^2+4x+4)+(-y^2+2y-1)
=(x+2)^2-(y-1)^2
=(x+y+1)(x-y+3)
9. 解:原式=x^4+4x^2-4x^4+4
=(x^2+2)^2-(2x)^2
=(x^2+2x+2)(x^2-2x+2)
10. 解:原式= a^2(b-c)+b^2c-ab^2+ac^2-bc^2
= a^2(b-c)-a(b^2-c^2)+(b^2c-bc^2)
= (b-c) [a^2 -a(b+c)+bc]
= (b-c)(a-b)(a-c)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询