That is, {lnx1= ax12+bx1lnx2 = ax22+bx2, and the result is: lnx1x2 = a (x1+x2). lnx 1x 2 =[a(x 1+x2)+b](x 1-x2)，

F ′ (x) =1x-2ax-b and 2x0=x 1+x2, f ′ (x0) =1x0-2ax0-b = 2x1+x2-65438+

= 1x 1-x2[2(x 1-x2)x 1+x2-lnx 1x 2]= 1x 1-x2[2(x 1x 2- 1)x 1x 2+ 1-lnx 1x 2]，

Let t = x 1x2 ∈ (0, 1) and φ (t) = 2t-2t+1-LNT (0 < t <1),

∵φ′(t)=-(t- 1)2t(t+ 1)2 < 0，

∴Φ (t) is a decreasing function on (0, 1),

∴φ(t)>φ( 1)=0，

And x 1 < x2,

∴f'(x0)