数学因式分解(分组分解)

2025-06-26 23:52:27
推荐回答(3个)
回答1:

解:1.a²b²-a²+2ab-b²
=(ab)²-(a²-2ab+b²)
=(ab)²-(a-b)²
=(ab+a-b)(ab-a+b)
2.x³-x²-x+1
=x³+1-x²-x
=(x+1)(x²-x+1)-x(x+1)
=(x+1)(x²-x+1-x)
=(x+1)(x-1)²
3. 9a²-b²-3a-b
=(3a+b)(3a-b)-(3a+b)
=(3a+b)(3a-b-1)
4.a²+2ab+b²-1
=(a+b)²-1
=(a+b+1)(a+b-1)
5 x²-y²-z²-2yz
=x²-(y²+2yz+z²)
=x²-(y+z)²
=(x+y+z)(x-y-z)
6.a²-4ab+4b²-4
=(a-2b)²-2²
=(a-2b+2)(a-2b-2)
7 4m²-x²-9-6x
=(2m)²-(x²+6x+9)
=(2m)²-(x+3)²
=(2m+x+3)(2m-x-3)
8.(x²-y²)+(6y-9)
=x²-(y²-6y+9)
=x²-(y-3)²
=(x+y-3)(x-y+3)
9.a²+2ab+b²-2a-2b+1
=(a+b)²-2(a+b)+1
=(a+b-1)²
希望对你帮助

回答2:

1,原式等于(ab)^2-(a-b)^2=(ab+a-b)(ab-a+b)
2,原式=x(x^2-1)-(x^2-1)=(x-1)(x^2-1)=(x-1)(x-1)(x+1)
3,原式=(3a-b)(3a+b)-(3a+b)=(3a+b)(3a-b-1)
4原式=(a+b)^2-1=(a+b+1)(a+b-1)
就这么多吧,原理都一样,平方差公式和完全平方公式的正逆运算

回答3:

解:1.原式 =(ab)^2-(a^2-2ab+b^2)
=(ab)^2-(a-b)^2
=(ab+a-b)(ab-a+b
2.原式=x(x^2-1)-(x^2-1)
=(x-1)(x^2-1)
=(x-1)(x+1)(x-1)
=(x-1)^2(x+1)
3.原式=(9a^2-b^2)-(3a+b)
=(3a+b)(3a-b)-(3a+b)
=(3a+b)(3a-b-1)
4.原式=(a+b)^2-1
=(a+b+1)(a+b-1)
5.原式 =x^2-(y^2+2yz+z^2)
=x^2-(y+z)^2
=(x+y+z)(x-y-z)
6.原式 =(a^2-4ab+4b^2)-4
=(a-2b)^2-2^2
=(a-2b+2)(a-2b-2)
7.原式=4m^2-x^2-9-6x
=4m^2-(x^2+9+6x )
=(2m)^2-(x+3)^2
=(2m+x+3)(2m-x-3)
8.原式= =x^2-(y^2-6y+9)
=x^2-(y-3)^2
=(x+y-3)(x-y+3)
9.原式=(a^2+2ab+b^2)-2(a+b)+1
=(a+b)^2-2(a+b)+1
=(a+b-1)^2