解方程[2(x^2+1)]⼀(x+1)+[6(x+1)]⼀(x^2+1)=7

2025-01-07 19:00:24
推荐回答(2个)
回答1:

令A=(X+1)/(X²+1)
∴原式得:
2/A+6A=7
6A²-7A+2=0
(3A-2)(2A-1)=0
3A-2=0 2A-1=0
∴A1=2/3 A2=1/2
A1=2/3时:
(x+1)/(x²+1)=2/3
3x+3=2x²+2
2x²-3x-1=0
x=3±√[(-3)²-4*(-1)*2]/4=(3±√17)/4
A2=1/2时
(x+1)/(x²+1)=1/2
x²+1=2x+2
x²-2x-1=0
x=2±√[(-2)²-4*(-1)]/2=1±√2
∴x1=(3±√17)/4
x2=1±√2

回答2:

设y=(x²+1)/(x+1)
2y+6/y=7
2y²-7y+6=0
(2y-3)(y-2)=0
∴y=3/2 y=2
即(x²+1)/(x+1)=3/2 (x²+1)/(x+1)=2
2x²+2=3x+3 x²+1=2x+2
2x²-3x-1=0 x²-2x-1=0
……

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