=1/(x-2)(x-1)+1/(x-2)(x-3)+1/(x-3)(x-4)-4/(x-1)(x-5) 幫你刪除了 1/x2-7
=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)-[1/(x-5)-1/(x-1)]
=1/(x-5)不對啊，為什么要把1/x2-7刪掉？你那個多寫了你仔細看一下我打錯了，sorry，原題是1/x2-3x+2+1/x2-5x+6+1/x2-7x+12+1/x2-9x+20-4/x2-6x+5再解一次好嗎？！上面我給你的答案就是這個的解不對，題目不一樣了，別人說正確答案是0，好嗎？題目不一樣了哦你里面還有一個1/x2-9x+20我加上去，剛才我本來少寫了一項 =1/(x-2)(x-1)+1/(x-2)(x-3)+1/(x-3)(x-4)+1(x-4)(x-5)-4/(x-1)(x-5) =1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)+1/(x-5)-1/(x-4)-[1/(x-5)-1/(x-1)] 第一項和第四項抵消第二項和最后一項抵消 =0