Question

几道因式分解题目

要有完整的过程

分解因式
8a^2-12ab^3c+6a^3b^2c

-x^3y^3+x^3y

-8ax^2+16axy-8ay^2

(a^2+1)^2-4a^2

m^2+2n-mn-2m

x2+6x-27

(x^2+5x+2)(x^2+5x+3)-12

(x+y)(x+y+2xy)+(xy+1)(xy-1)

4x^3y-12x^2y^2+9xy^3

3x^4y^2-15x^2y^2+12y^2

(x-y)^4(x+y)^2-10(x-y)^2(x+y)^2+9(x+y)^2

Answer 1

8a^2-12ab^3c+6a^3b^2c=2a(4a-6b^3c+3a^2b^2c)

-x^3y^3+x^3y=x^3y(1-y^2)=x^y(1+y)(1-y)

-8ax^2+16axy-8ay^2=-8a(x^2-2xy+y^2)=-8a(x-y)^2

(a^2+1)^2-4a^2=(a^2+1-2a)(a^2+1+2a)=(a-1)^2(a+1)^2

m^2+2n-mn-2m=m^2-mn-2m+2n=m(m-n)-2(m-n)=(m-2)(m-n)

x2+6x-27=(x+9)(x-3)

(x^2+5x+2)(x^2+5x+3)-12=(x^2+5x)^2+5(x^2+5x)-6=[x^2+5x+6][x^2+5x-1]=(x+2)(x+3)(x^2+5x-1)
=(x+2)(x+3)[x+(5+√29)/2][x+(5-√29)/2]

(x+y)(x+y+2xy)+(xy+1)(xy-1)=(x+y)^2+2xy(x+y)+(xy)^2-1=(x+y+xy)^2-1
=(x+y+xy+1)(x+y+xy-1)

4x^3y-12x^2y^2+9xy^3=xy(4x^2-12xy+9y^2)=xy(2x-3y)^2

3x^4y^2-15x^2y^2+12y^2=3y^2(x^4-5x^2+4)=3y^2(x^2-1)(x^2-4)=3y^2(x-1)(x+1)(x-2)(x+2)

(x-y)^4(x+y)^2-10(x-y)^2(x+y)^2+9(x+y)^2=(x+y)^2[(x-y)^4-10(x-y)^2+9]
=(x+y)^2[(x-y)^2-1][(x-y)^2-9]=(x+y)^2(x-y-1)(x-y+1)(x-y-3)(x-y+3)

Answer 2

8a^2-12ab^3c+6a^3b^2c=2a(4a-6b^3c+3a^2b^2c)

-x^3y^3+x^3y=x^3y(1-y^2)=x^3y(1+y)(1-y)

-8ax^2+16axy-8ay^2=-8a(x^2-2xy+y^2)=-8a(x-y)^2

(a^2+1)^2-4a^2=(a^2+1-2a)(a^2+1+2a)=(a-1)^2(a+1)^2

m^2+2n-mn-2m=m^2-mn-2m+2n=m(m-n)-2(m-n)=(m-2)(m-n)

x2+6x-27=(x+9)(x-3)

(x^2+5x+2)(x^2+5x+3)-12=(x^2+5x)^2+5(x^2+5x)-6=[x^2+5x+6][x^2+5x-1]=(x+2)(x+3)(x^2+5x-1)

(x+y)(x+y+2xy)+(xy+1)(xy-1)=(x+y)^2+2xy(x+y)+(xy)^2-1=(x+y+xy)^2-1
=(x+y+xy+1)(x+y+xy-1)

4x^3y-12x^2y^2+9xy^3=xy(4x^2-12xy+9y^2)=xy(2x-3y)^2

3x^4y^2-15x^2y^2+12y^2=3y^2(x^4-5x^2+4)=3y^2(x^2-1)(x^2-4)=3y^2(x-1)(x+1)(x-2)(x+2)

(x-y)^4(x+y)^2-10(x-y)^2(x+y)^2+9(x+y)^2=(x+y)^2[(x-y)^4-10(x-y)^2+9]
=(x+y)^2[(x-y)^2-1][(x-y)^2-9]=(x+y)^2(x-y-1)(x-y+1)(x-y-3)(x-y+3)

Answer 3

8a^2-12ab^3c+6a^3b^2c

-x^3y^3+x^3y

-8ax^2+16axy-8ay^2

(a^2+1)^2-4a^2

m^2+2n-mn-2m

x2+6x-27

(x^2+5x+2)(x^2+5x+3)-12

(x+y)(x+y+2xy)+(xy+1)(xy-1)

4x^3y-12x^2y^2+9xy^3

3x^4y^2-15x^2y^2+12y^2

(x-y)^4(x+y)^2-10(x-y)^2(x+y)^2+9(x+y)^2？