ï¼1ï¼ç±é¢ç®å¾ï¼f'(x)=[(ax^2+bx)'(x+1)-(x+1)'(ax^2+bx)]/(x+1)^2 å³f'(x)=(ax^2+2ax+b)/(x+1)^2
åå 为y=f(x)å¨x=1å¤çæç为ï¼ç±5x-4y+1=0å¾ï¼y=5/4x+1/4ï¼k=5/4
æ以ï¼f'(1)=(ax^2+2ax+b)/(x+1)^2=5/4 .解å¾ï¼3a+b=5
åæy=f(x)å¨ç¹(1,f(1))å¤çå线æ¹ç¨æ¯5x-4y+1=0ï¼å³f(1)=y=3/2
æ以ï¼f(1)=(ax^2+bx)/(x+1)=3/2 .解å¾ï¼a+b=3
æ
ï¼a=1ï¼b=2
(2)ãç±ï¼1ï¼å¾ï¼f(x)=(x^2+2x)/(x+1)ï¼æ以ï¼g(x)=2ln(x+1)-m(x^2+2x)/(x+1)
è¥å½xå±äº[0ï¼æ£æ ç©·ï¼æ¶ï¼ææg(x)<=0ï¼å³ï¼g(x)max<=0
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