2(x^2+1)/x+1 + 6(x+1)/x^2+1 =7
来源:百度知道 编辑:UC知道 时间:2024/05/30 17:33:03
2(x^2+1)/x+1与6(x+1)/x^2+1各为两个分式
要过程!!!!
要过程!!!!
2(x^2+1)/x+1 + 6(x+1)/x^2+1 =7
设:t=(x^2+1)/(x+1),则有:
2t+6/t-7=0
2t^2-7t+6=0
(2t-3)(t-2)=0
t1=3/2
t2=2
(1)t1=3/2
(x^2+1)/(x+1)=3/2
2x^2+2=3x+3
2x^2-3x-1=0
x=(3±√17)/4
(2)t2=2
(x^2+1)/(x+1)=2
x^2+1=2x+2
x^2-2x-1=0
x=1±√2
经检验:。。。。
解:设t=(x^+1)/(x+1),则有2t+6/t=7,所以2t^2-7t+6=0,解得t1=1/2,t2=3,当t=1/2时,(x^2+1)/(x+1)=1/2,此方程无实数根,当t=3时,,(x^2+1)/(x+1)=3,解得x1=1/2(3+根号17),x2=1/2(3-根号17)
经检验,x1=1/2(3+根号17),x2=1/2(3-根号17)是原方程的根
x=1(一眼就看出来了,真对不起~~~~~~~~~~~)
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