6、若 (x)=dfrac (1)(x-1), 则 f(x+1)= ()-|||-

[例26]已知 (dfrac (x+1)(x-1))=2f(x)-3x, 则 f(x)= __-|||-解:令 =dfrac (x+1)(x-1), 则 =df

[例26]已知 (dfrac (x+1)(x-1))=2f(x)-3x, 则 f(x)= __

1.设 (dfrac (1)(x))=x((dfrac {x)(x+1))}^2, 则 f(x)= ()-|||-(A) dfrac (1)(x)((dfrac

2 基础出 已知 (x)=dfrac (x-1)(x+1) ,求 f(-2) ,f[f(x)].

若f(x-1)=x^2(x-1)，则f(x)=(  )A. $$x(x+1)^2$$B. $$x(x-1)^2$$C. $$x^2(x+1)$$D. $$x^2

25 单选 (x)=x+(x-1)arcsin dfrac (x)(x+1) ,则 (1)=-|||-

将函数(x)=dfrac (1)(x+1)展开成(x-1)的幂级数2.计算(x)=dfrac (1)(x+1),其中(x)=dfrac (1)(x+1)是锥面(

极限 lim _(xarrow infty )((dfrac {x+1)(x-1))}^x= () .

极限 lim _(xarrow infty )((dfrac {x+1)(x-1))}^x= __

(16) lim _(xarrow infty )((dfrac {x-1)(x+1))}^x ;