解： ∵x^6+1 =(x^2+1)(x^4-x^2+1) =1/x*(x^2+1)(x^5-x^3+x) =2(x+1/x) 而x>0,且x不等于1, ∴x^6+1>4→x>3^(1/6). 故x∈(3^(1/6),+∞). 又,x^5-x^3+x=2 ∴x^3+2=x^5+x →1+2/x^3=x^2+1/x^2>2 →x^3<2,即x<2^(1/3). 故x∈(0,2^(1/3)). 综上知,原方程正实数根所属区间为 (0,2^(1/3))∪(3^(1/6),+∞).
答：解:x^6+1 =(x^2+1)(x^4-x^2+1) =1/x*(x^2+1)(x^5-x^3+x) =2(x+1/x) 因x>0且x不为1, 故x^6+1>...详情>>
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