chenlawer
幼苗
共回答了23个问题采纳率:91.3% 举报
第一问
令f(x)=ln(x+1)-(x-1/2 *x^2)
f'(x)=1/(x+1)-1+x
=x^2/(x+1)
x>0时f'(x)>0
f(x)单调增
f(0)=0
f(x)>f(0)=0
即 ln(x+1)> x-1/2*x^2
第二问
令g(x)=x-ln(x+1)
g'(x)=1-1/(x+1)
=x/(x+1)
x>0时g'(x)>0
g(x)单调增
g(0)=0
g(x)>g(0)=0
即 x> ln(x+1)
1年前
2