(2)证明: db, dB. 1 4b2N(-)b2+x2)22x 1 16 Nx(b+x) dx dx db dB dB, 三+a =0 (3)解: db, dB 01b2N(b2+x2)2+0b2Nx2(b2+x2) x 2 d2b d2B d B dx dx dx 15 b2N[d(b2+)2-3(b2+)2]（2）证明： 2 5 2 2 2 0 2 5 2 2 2 0 1 2 ( ) 2 3 )( ) 2 2 3 ( 2 1 − − = = Ib N − b + x x = − Ib N x b + x dx dB dx dB   0 2 2 2 1  = + = = =− d x d x dx dB dx dB dx dB (3)解： 2 7 2 2 2 2 0 2 5 2 2 2 2 0 2 2 2 1 2 ( ) 2 15 ( ) 2 3 − − = = − Ib N b + x + Ib N x b + x dx d B dx d B   2 2 2 1 2 2 2 2 2 2 2 2 7 5 2 2 2 2 2 2 0 1 15 [ ( ) 3( ) ] 2 4 4 4 d d x x d B d B d B dx dx dx d d  Ib N d b b = =− − − = + = + − +