證 (1)設f1(x),f2(x)均為偶函數(shù),令F(x)=f1(x)+f2(x),因F(-x)=f1(-x)+f2(-x)=f1(x)+f2(x)=F(x),故F(x)為偶函數(shù).設g1(x),g2(x)均為奇函數(shù),令G(x)=g1(x)+g2(x),因G(-x)=g1(-x)+g2(-x)=-g1(x)-g2(x)=-G(x),故G(x)為奇函數(shù).(2)設f1(x),f2(x)均為偶函數(shù),令F(x)=f1(x)·f2(x),因F(-x)=f1(-x)· (本文共 695 字 ) [閱讀本文] >>