(x²+x)²-(x²+x)-12=0用换元法解
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(x²+x)²-(x²+x)-12=0用换元法解
(x²+x)²-(x²+x)-12=0用换元法解
(x²+x)²-(x²+x)-12=0用换元法解
令x^2+x=t(t≥-1/4)
t^2-t-12=0
(t+3)(t-4)=0
t=-3(舍)或t=4
x^2+x-4=0
x=(-1±√17)/2