解誉正伏:当x>0时,-x<0f(-x)+f(x)=0,f(-x)=-x(1+x)∴f(x)=-f(-x)=x(1+x),清禅x>0x∈(0,1)时g(x)=f(x)-x+1/x=x²+1/xg'(x)=2x-1/x²g''(x)=2+2/x³>0g'(x)=0时,2x=1/x²,x³=1/2,x=(1/2)^(1/3)∴在(0,(1/庆携2)^(1/3))上g'(x)<0,g(x)递减,在((1/2)^(1/3),1)上g'(x)>0,g(x)递增,谢谢