f(x)=x^2(x-1)(x-2),则f'(x)的零点个数为顺便问问一元三次方程的根可不可能只有两非同根

若f(x)满足f(x)-2f(1/x)=x,则f(x)=? 若F(x)满足f(x)+2f(1/x)=x,则f(x)= 已知f(x+1/x)=x^2+1/x^2+1/x则f(x) f(x,-x)=x^2+2x,fx(x,-x)=5x+1,则fy(x,-x)= 已知f(x)+2f(1/x)=x.x不等于0.则f(x)= f(x)={2x+1,x f(x)={1+x/2,x f(x)=(x-1)/(x+2) f(x)=x^2+x (x 若函数f(x)满足f(x)+2f(1/x)=5x+4/x,则f(x)= 已知函数f(x)满足2f(x/1)-f(x)=x ,x不等于0,则f(x)等于 f(x)+f((x-1)/x)=2x; x!=0,1; 求f(x) f(x)+f[(x-1)/x]=2x x不等于0,1.求f(x). max{f(x),g(x)}=1/2(f(x)+g(x)+|f(x)-g(x)| f(x)满足f(-x)+2f(x)=x+3,则f(1)等于 f( (1-x)/(1+x) )=(1-x^2)/(1+x^2),则f(x)= 已知f(x-1/x+1)=x-2/x+2,则f(x)=. 已知f(x+1/x)=x^2+1/x^2,则f'(x)=