作业帮1.分解因式3x2-12-8a+20x22.化简1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)(a2+3a+3)/(a2+3a+2)-(a2+3a-2)(4-a2)+(1+2a-2a2)/(a2-4-2)
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1.分解因式3x2-12-8a+20x22.化简1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)(a2+3a+3)/(a2+3a+2)-(a2+3a-2)(4-a2)+(1+2a-2a2)/(a2-4-2)
1.分解因式
3x2-12-8a+20x2
2.化简
1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)
(a2+3a+3)/(a2+3a+2)-(a2+3a-2)(4-a2)+(1+2a-2a2)/(a2-4-2)
1.分解因式3x2-12-8a+20x22.化简1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)(a2+3a+3)/(a2+3a+2)-(a2+3a-2)(4-a2)+(1+2a-2a2)/(a2-4-2)
1、原式=2（3-6-4a+20）
=2（17-4a)
2、原式=1/x(x+1) + 1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4)
=[(x+2)+x]/x(x+1)(x+2) + [(x+4)+(x+2)]/(x+2)(x+3)(x+4)
=(2x+2)/x(x+1)(x+2) + (2x+6)/(x+2)(x+3)(x+4)
=2/x(x+2) + 2/(x+2)(x+4)
=[2(x+4)+2x]/x(x+2)(x+4)
=8/x(x+2)(x+4)
第3题的题目,第二部分两个括号中间应该有个"/",最后一个括号里的应该是不是(a^2-4)?
那么这题应该如下解.
原式=(a2+3a+3)/(a+1)(a+2) + (a2+3a-2)/(a+2)(a-2) +
(1+2a-2a2)/(a+2)(a-2)
=(a2+3a+3)/(a+1)(a+2)
+ [(a2+3a-2)+(1+2a-2a2)]/(a+2)(a-2)
=(a2+3a+3)/(a+1)(a+2) + (-a2+5a-1)/(a+2)(a-2)
=(a2+3a+3)(a-2)/(a+1)(a+2)(a-2)
+ (-a2+5a-1)(a+1)/(a+2)(a-2)(a+1)
=[(a3-2a2+3a2-6a+3a-6)+(-a3-a2+5a2+5a-a-1)]/(a+2)(a-2)(a+1)
=(5a2+a-7)/(a+2)(a-2)(a+1)
1.
原式=6-12-8a+40
=34-8a
=2(17-4a)