y²+1/y²+2y+(2-6y)/y=0
(y+1/y)²-2+2y+2/y-6=0
(y+1/y)²+2(y+1/y)-8=0
設(shè)y+1/y=t.則
t²+2t-8=0
(t+4)(t-2)=0
t=-4或t=2
(1)t=-4時(shí)
y+1/y=-4
y²+4y+1=0
(y+2)²=3
y1=-2+√3 y2=-2-√3
(2)t=2時(shí)
y+1/y=2
y²-2y+1=0
(y-1)²=1
y3=1
所以原方程的解為
y1=-2+√3 y2=-2-√3 y3=1