分母有一個三次方

可以參考寸絲老師的做法
https://math.pro/db/viewthread.php?tid=3507&page=4#pid22683

[ 本帖最後由 PDEMAN 於 2022-4-26 08:49 編輯 ]

將函數\(f(x)\div (x+1)(x-1)\)得到商式\((2x+a(a+3))\)

可以畫圖想一下題目的可能重根或大於1的根才有可能
所以滿足題目要求的條件需要第三根\((-\frac{a(a+3)}{2}\geq 1)\)

[ 本帖最後由 PDEMAN 於 2022-5-5 12:41 編輯 ]

打開絕對值應該可以分成四條線
1.\((a_{1}+a_{2})x+(b_{1}+b_{2})y+(c_{1}+c_{2}-1)=0\)
2.\((a_{1}-a_{2})x+(b_{1}-b_{2})y+(c_{1}-c_{2}-1)=0\)
3.\((a_{1}+a_{2})x+(b_{1}+b_{2})y+(c_{1}+c_{2}+1)=0\)
4.\((a_{1}-a_{2})x+(b_{1}+b_{2})y+(c_{1}-c_{2}+1)=0\)
1,3平行2,4,平行
用兩平行直線可算出距離為線段\(AB=\frac{2}{\sqrt{(a_{1}-a_{2})^2+(b_{1}-b_{2})^2}}=\frac{2}{\sqrt{(a_{1}+a_{2})^2+(b_{1}+b_{2})^2}}=5\)
推得\(\sqrt{(a_{1}-a_{2})^2+(b_{1}-b_{2})^2}=\frac{2}{5}\)和\(\sqrt{(a_{1}+a_{2})^2+(b_{1}+b_{2})^2}=\frac{2}{5}\)
又1,3或2,4與直線\(AB:y-3=-\frac{3}{4}(x-6)\)為同一條
所以1,3或2,4為直線\(AB:y-3=-\frac{3}{4}(x-6)\)的\(\frac{2}{25}\)倍
可以知道\(c_{1}-c_{2}+1 ,c_{1}-c_{2}+1,c_{1}-c_{2}-1,c_{1}+c_{2}-1\)再取大的即可

[ 本帖最後由 PDEMAN 於 2022-5-6 07:57 編輯 ]