请问“(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)”怎样做,急用

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 06:51:21
请问“(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)”怎样做,急用

请问“(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)”怎样做,急用
请问“(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)”怎样做,急用

请问“(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)”怎样做,急用
分子都比分母大1,所以先整理为: 原式=(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)] =1+1/(x+1)-1-1/(x+2)-1+1/(x-3)+1-1/(x-4) =[1/(x+1)+1/(x-3)]-[1/(x+2)+1/(x-4)] =[(x-3)+(x+1)]/[(x+1)(x-3)]-[(x-4)+(x+2)]/[(x+2)(x-4)] =(2x-2)/(x^2-2x-3)-(2x-2)/(x^2-2x-8) =(2x-2)*[1/(x^2-2x-3)-1/(x^2-2x-8)] =(2x-2)*[(x^2-2x-8)-(x^2-2x-3)]/[(x^2-2x-3)(x^2-2x-8)] =(2x-2)*(-5)/[(x^2-2x-3)(x^2-2x-8)] =(10-10x)/[(x^2-2x-3)(x^2-2x-8)]
满意请采纳