解方程:(x²+x+1)/(x²+1)+(2x²+x+2)/(x²+x+1)=1
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/04/29 12:14:19
解方程:(x²+x+1)/(x²+1)+(2x²+x+2)/(x²+x+1)=19/6
:(x²+x+1)/(x²+1)+(2x²+x+2)/(x²+x+1)=19/6
:(x²+x+1)/(x²+1)+(x²+x+1+x²+1)/(x²+x+1)=19/6
:(x²+x+1)/(x²+1)+1+ (x²+1)/(x²+x+1)=19/6
设:(x²+x+1)/(x²+1)=y
∴原方程可化为
y +1+1/y=19/6
6y²-13y+6=0
(3y-2)(2y-3)=0
∴y=2/3 y=3/2
∴:(x²+x+1)/(x²+1)=2/3 :(x²+x+1)/(x²+1)=3/2
2x²+2=3x²+3x+3 3x²+3=2x²+2x+2
x²+3x+1=0 x²-2x+1=0
x=(-3±√5)/2 x=1
:(x²+x+1)/(x²+1)+(x²+x+1+x²+1)/(x²+x+1)=19/6
:(x²+x+1)/(x²+1)+1+ (x²+1)/(x²+x+1)=19/6
设:(x²+x+1)/(x²+1)=y
∴原方程可化为
y +1+1/y=19/6
6y²-13y+6=0
(3y-2)(2y-3)=0
∴y=2/3 y=3/2
∴:(x²+x+1)/(x²+1)=2/3 :(x²+x+1)/(x²+1)=3/2
2x²+2=3x²+3x+3 3x²+3=2x²+2x+2
x²+3x+1=0 x²-2x+1=0
x=(-3±√5)/2 x=1
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