行列式1 -1 1 x-1-|||-1 -1 x+1 -1-|||-1 x-1 -1 0-|||-x+1 -1 0 0的值为1 -1 1 x-1-|||-1 -1 x+1 -1-|||-1 x-1 -1 0-|||-x+1 -1 0 0。1 -1 1 x-1-|||-1 -1 x+1 -1-|||-1 x-1 -1 0-|||-x+1 -1 0 01 -1 1 x-1-|||-1 -1 x+1 -1-|||-1 x-1 -1 0-|||-x+1 -1 0 01 -1 1 x-1-|||-1 -1 x+1 -1-|||-1 x-1 -1 0-|||-x+1 -1 0 01 -1 1 x-1-|||-1 -1 x+1 -1-|||-1 x-1 -1 0-|||-x+1 -1 0 0

5.求解下列方程:-|||-x+1 2 -1-|||-(1) 2 x+1 1 =0;-|||--1 1 x+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(

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

5.求解下列方程:-|||-x+1 2 -1-|||-(1) 2 x+1 1 =0;(2)-|||--1 1 x+1-|||-互不相等.-|||-=0,其中a,

(1) =dfrac (x+1)(x(x-1)) 在区间 (-1,0) (0,1)、(1,2)、(2,3)上的有界性;

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

设A=1 -1 0-|||-0 1 -1-|||--1 0 1-|||-__，AX=2X+A,求X.设A=，AX=2X+A,求X.

2.试用拉普拉斯定理计算行列式-|||-1 1 1 0 0-|||-1 2 3 0 0-|||-D= 0 1 1 1 1-|||-0 x1 x2 x3 x4-|

【65】int(x^2+1)/((x+1)^2)(x-1)dx.【65】$\int\frac{x^{2}+1}{(x+1)^{2}(x-1)}dx$.

(2) lim _(xarrow infty )((dfrac {x+1)(x-1))}^-dfrac (x{2)}= () ;-|||-(A)1 (B) ^d