解分式方程1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)1/(x²+9x+20)=1/8我不知道是解错了还是怎么回事,算到最后式子中含有x²项.

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/08 03:09:30

解分式方程1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)1/(x²+9x+20)=1/8我不知道是解错了还是怎么回事,算到最后式子中含有x²项.
解分式方程1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)1/(x²+9x+20)=1/8
我不知道是解错了还是怎么回事,算到最后式子中含有x²项.

解分式方程1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)1/(x²+9x+20)=1/8我不知道是解错了还是怎么回事,算到最后式子中含有x²项.
1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)+1/(x²+9x+20)=1/8
1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/8
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)=1/8
1/(x+1)-1/(x+5)=1/8
8(x+5)-8(x+1)=(x+1)(x+5)
8x+40-8x-8=x²+6x+5
x²+6x-27=0
(x+9)(x-3)=0
x=-9或x=3
经检验x=-9或x=3都是方程的根