(1)证:若f(x)≠0,则 x(g2(x)+h2(x)=f(x)≠0 从而g2(x)+h2(x)≠0.于是 (xg2(x)+xh2(x)=(x(g2(x)+h2(x)为奇数 但a(f2(x))为偶数∴x(g2(x)+h2(x)≠f2(x) 这与已知矛盾.故∫(x)=0, 从而82(x)+h(x)=0.(1) 证：若 f x( ) 0,  则 2 2 2 x g x h x f x ( ( ) ( )) ( ) 0, + =  于是 2 2 2 2  + =  + ( ( ) ( )) ( ( ( ) ( ))) xg x xh x x g x h x 为奇数. 故 f x( ) 0, = 从而 2 2 g x h x ( ) ( ) 0. + = 从而 2 2 g x h x ( ) ( ) 0. +  2 但 ( ( )) f x 为偶数. 这与已知矛盾. 2 2 2  +  x g x h x f x ( ( ) ( )) ( )