[例26]已知 (dfrac (x+1)(x-1))=2f(x)-3x, 则 f(x)= __-|||-解:令 =dfrac (x+1)(x-1), 则 =dfrac (t+1)(t-1), 得到-|||-(t)=2f(dfrac (t+1)(t-1))-dfrac (3t+3)(t-1)=2[ 2f(t)-3t] -dfrac (3t+3)(t-1),-|||-从而, (t)=2t+dfrac (t+1)(t-1), 因此, (x)=2x+dfrac (x+1)(x-1).

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

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

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)].

设(x)+(x)^2f(dfrac (1)(x))=dfrac ({x)^2+2x}(x+1) 求f(x)设

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(

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

设函数=dfrac ({x)^2}(x-1),则=dfrac ({x)^2}(x-1)=________.设函数,则=________.

int dfrac ({x)^2+1}({(x+1))^2(x-1)}dx；；

[例8]设f(x )在 x=1 连续,且 lim _(xarrow 1)dfrac (f(x)-2)(x-1) 存在,则 (1)= __-|||-解:因为 li