[ dfrac (sin z)({z)^2},0] =-|||-A 1-|||-B .-1-|||-C dfrac (1)(2)

【题目】 (int )_(0)^dfrac (pi {2)}cos xdx= ()-|||-A、 dfrac (1)(2)-|||-B、1-|||-C、 -df

4.设 (z)=dfrac (1)(z)-zsin dfrac (1)({z)^2}, 则 [ f(z),0] =-|||-(A)1; (B)2; (C)0;

-1 0 -1 0 0 的值为 ()-|||-A 1-|||-B ((-1))^dfrac (n(n-1){2)}-|||-C -1-|||-D ((-1))

计算(int )_(c)_(c)dfrac (cos z)((z-dfrac {1)(2))(z-1)}dz，其中(int )_(c)_(c)dfrac (co

曲线=dfrac (x)(1-{x)^2}的渐近线是（ ）.=dfrac (x)(1-{x)^2}和=dfrac (x)(1-{x)^2}=dfr

z=0是函数(z)=dfrac (z)(sin {z)^2cdot ((e)^z-1)}的几级极点A 1 B 2 C 3 D 4z=0是函数的

24、单选-|||-设 -|||-A 2-|||-B dfrac (1)(2)-|||-C 1-|||-D 0

=dfrac (arcsin x)(x)+dfrac (1)(2)ln dfrac (1-sqrt {1-{x)^2}}(1+sqrt {1-{x)^2}}

(z)=(z)^2+dfrac (1)({z)^2+1}，则其解析区域为( )(z)=(z)^2+dfrac (1)({z)^2+1}(z)=(z)^2+dfr

(z)=(z)^2+dfrac (1)({z)^2-1}，则其解析区域为（）(z)=(z)^2+dfrac (1)({z)^2-1}(z)=(z)^2+dfra