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f'(x)=ln(x+1)+1+2m(x+1)=ln(x+1)+2mx+2m+1已知x≥0則ln(x+1)≥ln1=0又f(x)≤0恒成立則必需f'(x)≤0所以2mx+2m+1≤0即m≤-1/2(x+1)≤-1/2此時,函數單减只需f(0)=0≤0故m≤-1/2希望能幫到你O(∩_∩)O…
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