1）.1/x²-3x+2 + 1/x²-5x+6 + 1/x²-4x+3

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/11/06 12:42:29

1）.1/x²-3x+2 + 1/x²-5x+6 + 1/x²-4x+3
2）.x-1/x²+3x+2 + 6/2+x-x² -(10-x)/4-x²
3）.（x+1-1/1-x）÷（x-x²/x-1）
4）.（x²-4/x²-x-6 +x+2/x-3）÷（x+1）/（x-3）
5）.x/x-y ·y²/x+y - x^4y/x^4-y^4 ÷ x²/x²+y²
"/" 分数线 .

不知道你的要求是不是化简,而且很多地方表达的有歧义性.我按化简这个方向和自己猜的提意做,你看对否?
1) [1/(x²-3x+2)] + [1/(x²-5x+6)] + [1/(x²-4x+3)]
原式=[1/(x-1)(x-2)]+[1/(x-2)(x-3)]+[1/(x-1)(x-3)]
=[(x-3)/(x-1)(x-2)(x-3)]+[(x-1)/(x-1)(x-2)(x-3)]+[(x-2)/(x-1)(x-2)(x-3)]
=[x-3+x-2+x-1]/[(x-1)(x-2)(x-3)]
=[3x-6]/[(x-1)(x-2)(x-3)]
=3/[(x-1)(x-3)]
2) [(x-1)/(x²+3x+2)] + [6/(2+x-x²)] -[(10-x)/(4-x²)]
原式=[(x-1)/(x+1)(x+2)]+[6/(x+1)(2-x)]-[(10-x)/(2-x)(2+x)]
=[(x-1)(2-x)/(x+1)(x+2)(2-x)]+[6(x+2)/(x+1)(x+2)(2-X)]-[(10-x)(1+x)/(1+x)(2+x)(2-x)]
=0/ [(1+x)(2+x)(2-x)]
=0
3) [（x+1)-1/(1-x）]÷[（x)-x²/(x-1）]
原式= [(x+1)(1-x)/(1-x)]-[1/(1-x)]÷[x(x-1)/(x-1)]-[x²/(x-1)]
=[(-x²)/(1-x)]÷[(-x)/(x-1)]
=[(-x²)/(1-x)]×[(1-x)/x]
=-x
4) [（x²-4)/(x²-x-6) +(x+2)/(x-3)]÷[（x+1）/（x-3）]
原式=[(x-2)(x+2)/(x-3)(x+2)+(x+2)/(x-3)]÷[(x+1)/(x-3)]
=[(x-2)/(x-3)+(x+2)/(x-3)]×[(x-3)/(x+1)]
=[2x/(x-3)]×[(x-3)/(x+1)]
=2x/(x+1)
5) x/(x-y) ·y²/(x+y) - x^4y/(x^4-y^4) ÷ x²/(x²+y²)
原式=xy²/(x²-y²)-x^4y/(x²+y²)(x²-y²)×(x²+y²)/x²
=xy²/(x²-y²)-x²y/(x²-y²)
=(xy²-x²y)/(x²-y²)
=xy(y-x)/(x+y)(x-y)
=-xy/(x+y)
PS 麻烦写的规范点实在是看的累

（x²-5x+6分之x²-3x+2）²/（x²-2x-3分之x-1）² 化简~(x²+3x+9/x²-27)+(6x/9x-x²)-(x-1/6+2x) 已知关于x的多项式（2mx²-x²+3x+1）-(5x²-4y²+3x)不含x& 2(x²-1²) 7/x²+x+3/x²-x=6/x²-1 已知：多项式（2mx²-x²+3x-1）-（5x²-4y²+3x²）不 1/X²+3X+2 +1/X²+5X+6 +1/X²+7X+12 +1/X²+9 17²-(8x²+5x)-(4x²+x-3)+(-5x²+6x-1)-3,其中x 试说明不论X取何值代数式（x²+5x²+4x-1)-(-x²-3x+2x²-3) 再求代数式（4x²-7x+6y-1）-（-2x²+3y²+2y+2x）+（-6x² （2mx²-x²+3x+1）-（5x²+4y²+3x）的值与x无关,求2m&su 多项式（2mx²-x²+3x+1)-(15x²-4x²+3x)不含X²