我說的詳細(xì)點(diǎn)：∵f(x)=(x^2 +1/(bx+c)為奇函數(shù) ∴有f(-x)=-f(x)--------------------① ∴由①可建立等式：(-x)^2 +1/(b(-x)+c)=-(x^2 +1/(bx+c) 得：-c=c,∴c=0 ∴f(x)=(x^2 +1/bx) 又∵f(1)=2 ∴2/b=2∴b=1 ∴f(x)=(x^2 /x)=x+1/x 設(shè)x1,x2且x1＜x2 ∴f(x2)-f(x1)=x2+1/x2-x1-1/x1 =(x2-x1)+(1/x2-1/x1) =(x2-x1)+(x1-x2)/x1x2 =(x2-x1)-(x2-x1)/x1x2 =(x2-x1)(x1x2-1/x1x2) ∴當(dāng)X屬于[-1,0)或（0,1]時(shí),X1X2-1＜0 ∵x2-x1＞0,x1x2＞0,X1X2-1＜0 ∴(x2-x1)(x1x2-1/x1x2)＜0,即f(x2)-f(x1)＜0,f(x2)＜f(x1) ∴當(dāng)X屬于[-1,0)或（0,1]時(shí),f(x)的單調(diào)區(qū)間為遞減 又當(dāng)X屬于(-∞,-1)或(1,∞)時(shí),X1X2-1＞0 ∵x2-x1＞0,x1x2＞0,X1X2-1＞0 ∴(x2-x1)(x1x2-1/x1x2)＞0,即f(x2)-f(x1)＞0,f(x2)＞f(x1) ∴當(dāng)X屬于(-∞,-1)或(1,∞)時(shí),f(x)的單調(diào)區(qū)間為遞增 哪里不明白可以再問---------------------------------------------------詠然不悔