分解因式::1.(x²+5x+3)(x²+5x-2)-6 2.(x+1)(x+3)(x+5)(x+7)-9 3. 5x²-21x&su

2025-01-01 22:28:04
推荐回答(5个)
回答1:

1.(x²+5x+3)(x²+5x-2)-6
=(x²+5x)²+3(x²+5x)-2(x²+5x)-6
=(x²+5x)²+(x²+5x)-6
=(x²+5x+3)(x²+5x-2)
2.(x+1)(x+3)(x+5)(x+7)-9
=.(x+1)(x+7)(x+3)(x+5)-9
=(x²+8x+7)(x²+8x+15)-9
=(x²+8x)²+22(x²+8x)+96
=(x²+8x+16)(x²+8x+6)
=(x+4)²(x²+8x+6)
3 5x³-21x²+18x
=x(5x²-21x+18)
=x(5x-6)(x-3)

回答2:

1..(x²+5x+3)(x²+5x-2)-6=(x²+5x)(x²+5x)+(x²+5x)-12
=(x²+5x+4)(x²+5x-3)
=(x+1)(x+4)(x²+5x-3)
2.(x+1)(x+3)(x+5)(x+7)-9
=(x+1)(x+7)(x+3)(x+5)-9
=(x²+8x+7)(x²+8x+15)-9
=(x²+8x)(x²+8x)+22(x²+8x)+96
=(x²+8x+16)(x²+8x+6)
=(x+4)²(x²+8x+6)
3.5x³-21x²+18x
=x(5x²-21x+18)
=x(5x-6)(x-3)

回答3:

1)设x^2+5x=y,有原式=(y+3)(y-2)-6=y^2+y-12=(y+4)(x-3)=(x^2+5x+4)(x^2+5x-3)=(x+1)(x+4)(x^2+5x-3) 2)设x^2+8x=y,原式=(y+7)(y+15)-9=(y+16)(y+6)=(x+4)(x+4)(x^2+8x+6)

回答4:

(x²+5x+3)(x²+5x-2)-6
=(x²+5x)²+(x²+5x)-6-6
=(x²+5x)²+(x²+5x)-12
=(x²+5x-3)(x²+5x+4)
=(x²+5x-3)(x+4)(x+1)
(x+1)(x+3)(x+5)(x+7)-9
=(x+1)(x+7)(x+3)(x+5)-9
=(x²+8x+7)(x²+8x+15)-9
=(x²+8x)²+22(x²+8x)+105-9
=(x²+8x)²+22(x²+8x)+96
=(x²+8x+16)(x²+8x+6)
=(x+4)²(x²+8x+6)
5x³-21x²+18x
=x(5x²-21x+18)
=x(5x-6)(x-3)

回答5:

1.(x²+5x+3)(x²+5x-2)-6
=(x^2+5x)^2+(x^2+5x)-12
=(x^2+5x+4)(x^2+5x-3)
=(x+4)(x+1)(x^2+5x-3)
2.(x+1)(x+3)(x+5)(x+7)-9
=(x^2+8x+7)(x^2+8x+15)-9
=(x^2+8x)^2+22(x^2+8x)+96
=(x^2+8x+16)(x^2+8x+6)
=(x+4)^2(x^2+8x+6)
3. 看不清