1).1/x²-3x+2 + 1/x²-5x+6 + 1/x²-4x+32).x-1/x²+3x+2 + 6/2+x-x² -(10-x)/4-x²3).(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

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 04:07:52
1).1/x²-3x+2 + 1/x²-5x+6 + 1/x²-4x+32).x-1/x²+3x+2 + 6/2+x-x² -(10-x)/4-x²3).(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

1).1/x²-3x+2 + 1/x²-5x+6 + 1/x²-4x+32).x-1/x²+3x+2 + 6/2+x-x² -(10-x)/4-x²3).(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
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 .

1).1/x²-3x+2 + 1/x²-5x+6 + 1/x²-4x+32).x-1/x²+3x+2 + 6/2+x-x² -(10-x)/4-x²3).(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
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