高校问答星际争霸2教学:数学题目
来源：百度文库 编辑：高校问答 时间：2024/04/19 18:24:34
解方程(1/(x^2+7x))-(1/(x^2+7x+12))+(1/(x^2+7x+18))-(1/(x^2+7x+12)=0
提示:设x^2+7x+9=y,原方程化为:(1/(y-9))-(1/(y-3))+(1/(y+9))-(1/(y+3))=0
提示:设x^2+7x+9=y,原方程化为:(1/(y-9))-(1/(y-3))+(1/(y+9))-(1/(y+3))=0
很简单!但有些麻烦,过程很多,我尽量说细些
设x^2+7x+9=y,则x^2+7x=y-9
原方程化为:(1/(y-9))-(1/(y+3))+(1/(y+9))-(1/(y+3))=0
通分后得(y+9)(y+3)+(y-9)(y+3)-2(y+9)(y-9)=0
打开括号整理得:y=27
把y=27代回x^2+7x+9=y
解一元二次方程得x=2;x=-9
代回原方程验证，x=2;x=-9均是原方程的解
所以原方程的根为x=2;x=-9
1/(x^2+7x+12) 且 x^2+7x+9=y 换成 1/(y-3) 这是怎么换的
因该是1/(y+3)吧
那就是原方程化为:(1/(y-9))-(1/(y+3))+(1/(y+9))-(1/(y+3))=0
通分的(y+9)(y+3)+(y-9)(y+3)-2(y+9)(y-9)=0
解的:y=27 接下来自己算吧
1/(x^2+7x+12) 且 x^2+7x+9=y 换成 1/(y-3) 这是怎么换的
因该是1/(y+3)吧
那就是原方程化为:(1/(y-9))-(1/(y+3))+(1/(y+9))-(1/(y+3))=0
通分的(y+9)(y+3)+(y-9)(y+3)-2(y+9)(y-9)=0
解的:y=27 接下来自己算吧
1/(x^2+7x+12）是不是抄错了，如果按:(1/(y-9))-(1/(y-3))+(1/(y+9))-(1/(y+3))=0计算，y=0.x为无理数自己算吧