设x' = x + 1 => x = x' - 1 这个式子成立吗?

设f(x)=1/x(x 设f(x)=1-x,(x 设f(x)=(x平方-1)/(x-1) (x 设f(x)=(x平方-1)/(x-1) (x 设f(x)=x/2+1/x(x 设函数f(x)={a/x+b/(x²-x) x>1{x x 设U={x|x 设f(x)={█( x ,x 设f(x)=x2+x(x 设f(x)=x(x 设f(x-1)={-sinx/x,x>0;2,x=0;x-1,x 设f(x-1)={-sinx/x,x>0;2,x=0;x-1,x 设2f(x)+xf(1/x)=(x+2x)/(x+1),求f(x). 设函数f(x)=(x-1)(x-2)...(x-100)(x>100),求F'(X) 设f（x）={x^sin(1/x),x>0 a+e^x,x 设函数f(x)=x-[x],x≥0,f(x+1),x 设f(x)满足f(1/x)=x/1-x设设f(x)满足f(1/x)=x/1-x ,则f(x+1)=? 设f(x)={1+x (-1