( 1)? What is the deformation of tangent equation? y =(- 1/2)(x- 1)+ 1

Visible slope k=- 1/2,? f( 1)= 1

f'(x)=[a(x+ 1)/x-alnx]/(x+ 1)^2-b/x^2

Known k = f' (1) = (2a)/4-b =-1/2? That's a-2b=- 1? ( 1)

f( 1)=b= 1？

Substitute (1) to get? a= 1

(2)? When x & gt0f (x) = lnx/(x+1)+1/x.

f(x)-lnx/(x+ 1)-k/x =( 1-k)/x & gt； 0

As long as1-k > 0 is enough.

So k < 1

However, (1)? What is the deformation of tangent equation? y =(- 1/2)(x- 1)+ 1

Visible slope k=- 1/2,? f( 1)= 1

f'(x)=[a(x+ 1)/x-alnx]/(x+ 1)^2-b/x^2

Known k = f' (1) = (2a)/4-b =-1/2? That's a-2b=- 1? ( 1)

f( 1)=b= 1？

Substitute (1) to get? a= 1

(2)? When x & gt0f (x) = lnx/(x+1)+1/x.

f(x)-lnx/(x+ 1)-k/x =( 1-k)/x & gt； 0

This method is the same as yours. However, there are other ways.

That's it.