10x²+30x+20约分 怎么会变成10（x+1）（x+2） x³+2x²-x-2约分怎么变成（x+2）(x+1) (x-1)把过程写清楚些?

10x²+30x+20约分 怎么会变成10（x+1）（x+2） x³+2x²-x-2约分怎么变成（x+2）(x+1) (x-1)把过程写清楚些?
10x²+30x+20约分 怎么会变成10（x+1）（x+2） x³+2x²-x-2约分怎么变成（x+2）(x+1) (x-1)
把过程写清楚些?

10x²+30x+20约分 怎么会变成10（x+1）（x+2） x³+2x²-x-2约分怎么变成（x+2）(x+1) (x-1)把过程写清楚些?
不是约分,是因式分解.
10x^2+30x+20=10(x^2+3x+2) (提取公因式）
=10（x+1)(x+2) (十字相乘法）.
x^3+2x^2--x--2=(x^3+2x^2)--(x+2) (分组）
=x^2(x+2)--(x+2) (前面一组提取公因式）
=(x+2)(x^2--1) (提取公因式）
=（x+2)(x+1)(x--1) (运用平方差公式）.

1. 10x²+30x+20=10(x²+3x+2)=10[(x²+x)+(2x+2)]=10[x(x+1)+2(x+1)]=10(x+2)(x+1)
2. x³+2x²-x-2=(x³+2x²)-(x+2)=x²(x+2)-(x+2)=(x+2)(x²-1)=(x+2)(x+1)(x-1)

10x²+30x+20
=10(x²+3x+2)
= 10 (x²+x+2x+2)
= 10 {(x²+x)+(2x+2)}
= 10 {x(x+1)+2(x+1)}
= 10 (x+1)(x+2)
x³+2x²-x-2
= (x³+2x²)-(x+2)
...

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10x²+30x+20
=10(x²+3x+2)
= 10 (x²+x+2x+2)
= 10 {(x²+x)+(2x+2)}
= 10 {x(x+1)+2(x+1)}
= 10 (x+1)(x+2)
x³+2x²-x-2
= (x³+2x²)-(x+2)
= x²(x+2)-(x+2)
= (x+2)(x²-1)
= (x+2)(x²+x-x-1)
= (x+2){(x²+x)-(x+1)}
= (x+2){x(x+1)-(x+1)}
= (x+2)(x+1)(x-1)

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10x²+30x+20
=10(x²+3x+2)
=10(x+1)(x+2)
十字相乘法
x 1
x 2
x³+2x²-x-2
=x²(x+2)-(x+2)
=(x+2)(x²-1)
=(x+2)(x+1)(x-1)=(x+2)(x^2--1) 这个...

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10x²+30x+20
=10(x²+3x+2)
=10(x+1)(x+2)
十字相乘法
x 1
x 2
x³+2x²-x-2
=x²(x+2)-(x+2)
=(x+2)(x²-1)
=(x+2)(x+1)(x-1)

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约分题：
(9x^2-6x+1)/(3x^2-x)
=[(3x-1)^2]/[x(3x-1)]
=(3x -1)/x
(4x^2-2x)/(4x^2-1)
=[2x(2x-1)]/[(2x+1)(2x-1)]
=(2x)/(2x+1)
(x^2-10x+25)/(3x^2-75)
=(x-5)^2/[3(x^2-25)]
=(...

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约分题：
(9x^2-6x+1)/(3x^2-x)
=[(3x-1)^2]/[x(3x-1)]
=(3x -1)/x
(4x^2-2x)/(4x^2-1)
=[2x(2x-1)]/[(2x+1)(2x-1)]
=(2x)/(2x+1)
(x^2-10x+25)/(3x^2-75)
=(x-5)^2/[3(x^2-25)]
=(x-5)^2/[3(x-5)(x+5)]
=(x-5)/[3(x+5)]
(2a^2-8)/(a^2+4a+4)
=[2(a^2-4)]/(a+2)^2
=[2(a-2)(a+2)]/(a+2)^2
=[2(a-2)]/(a+2)
[-3(a-b)]/[12(b-a)^2]
=-3/[12(a-b)]
=- 1/4(a-b)
(3x^2-12)/(x^2-4x+4)
=[3(x^2-4)]/(x-2)^2
=[3(x-2)(x+2)]/(x-2)^2
=[3(x+2)]/(x-2)
(x^2-3x)/(-x^2+6x-9)
=[x(x-3)]/[-(x-3)^2]
=x/[-(x-3)]
=- x/(x-3)
(2a^2-8ab+8b^2)/(2a^2-8b^2)
=[2(a^2-4ab+4b^2)]/[2(a^2-4b^2)]
=[2(a-2b)^2]/[2(a+2b)(a-2b)]
=(a-2b)/(a+2b)
(4b-10a)/(25a^2-20ab+4b^2)
=[2(2b-5a)]/(5a-2b)^2
=2/(2b-5a)
计算题：
[(3a^2b)/(2a-4b)]·[(a^2-4b^2)/(6ab)]
=[(3a^2b)(a^2-4b^2)]/[(2a-4b)(6ab)]
=[(3a^2b)(a-2b)(a+2b)]/[2(a-2b)(6ab)]
=[(3a^2b)(a+2b)]/(12ab)]
=[a(a+2b)]/4
[(x^2-2x+1)/(2x^2-18)]÷[(3-3x)/(4x^2-12x)]
=[(x^2-2x+1)/(2x^2-18)]·[(4x^2-12x)/(3-3x)]
=[(x^2-2x+1)(4x^2-12x)]/[(2x^2-18)(3-3x)]
=[(x-1)^2*4x(x-3)]/[6(x^2-9)(1-x)]
=[(x-1)^2*4x(x-3)]/[6(x-3)(x+3)(1-x)]
=[(1-x)*4x]/[6(x+3)]
=[2x(1-x)]/[3(x+3)]
7月e0

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8ab^(1-x)分之4ab(x-1