1.计算 (x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)2.解方程2x/

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/04/30 15:16:40

1.计算
(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
2.解方程
2x/(2x-1)+x/(x-2)=2
3.先化简再求值求(1/x²-1/y²)[(x²-xy+y²)/(x-y)+(x²+xy+y²)/(x+y)]的值 ,其中x=1/2 y=1/4

第一个：
(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
=1+1/(x+1) - 1 - 1/(x+2) - 1 +1/(x-3) +1 -1/(x-4)
= (x+2 -x -1)/[(x+1)*(x+2)] +(x-4 -x +3)/[(x-3)*(x-4)]
=1/[(x+1)*(x+2)] +(-1)/[(x-3)*(x-4)]
=[(x-3)*(x-4) - (x+1)*(x+2)] / [(x+1)*(x+2)*(x-3)*(x-4)]
=(10-10x) / [(x+1)*(x+2)*(x-3)*(x-4)]
第二问：
2x/(2x-1)+x/(x-2)=2
1+1/(2x-1) + 1+2/(x-2) = 2
1/(2x-1) + 2/(x-2) =0
1/(2x-1) = -2/(x-2)
2x-1=(-1/2)(x-2)
4x -2 =2 -x
5x=4
x=4/5
第三问：
(1/x²-1/y²) * [(x²-xy+y²)/(x-y)+(x²+xy+y²)/(x+y)]
=(1/x²-1/y²) * {(x+y)*(x²-xy+y²)/[(x-y)*(x+y)] + (x-y)*(x²+xy+y²)/[(x-y)*(x+y)]}
=(1/x²-1/y²) * [ (x³+y³)/(x² - y²) + (x³ -y³)/(x² - y²) ]
=[(1/x +1/y)*(1/x - 1/y) ] *[ (x³+y³)/(x² - y²) + (x³ -y³)/(x² - y²) ]
=[ (y²-x²)/(xy) ] * [ (x³+y³)/(x² - y²) + (x³ -y³)/(x² - y²) ]
=[ (y²-x²)/(xy) ] * 2x³/(x² - y²)
=-2x²/y
代入x=1/2 y=1/4
得到
-2x²/y
=-2 * (1/2)²/(1/4)
=-2 * (1/4) * 4
= -2
所以原式(1/x²-1/y²)[(x²-xy+y²)/(x-y)+(x²+xy+y²)/(x+y)] =-2

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