5.设 ,b,c,mu gt 0 ,曲面 =mu 与曲面 dfrac ({x)^2}({a)^2}+dfrac ({y)^2}({b)^2}+dfrac ({z)^2}({c)^2}=1 相切,则 μ= __

(2)已知 =V(x,y,z)=dfrac ({x)^2}({a)^2}+dfrac ({y)^2}({b)^2}+dfrac ({z)^2}({c)^2},

(B) (mu ,dfrac (1)(sqrt {2pi )})-|||-(C) (mu ,dfrac (1)(2)), (D)(0,σ).

下列选项中曲面dfrac ({x)^2}(4)+dfrac ({y)^2}(1)+dfrac ({z)^2}(9)=3上点dfrac ({x)^2}(4)+df

i 1设曲面 :dfrac ({x)^2}(2)+(y)^2+dfrac ({z)^2}(4)=1 及平面 pi :2x+2y+z+5=0.-|||-(1)求曲

3、设 (x,y)=arctan dfrac (x)(y), 则 (1,1)=-|||-(A)1; (B)0; (C) dfrac {1)(2),dfrac

(5)设 (x,y)=ln (x+dfrac (y)(2x)) ,则 _(y)(1,0)= () .-|||-(A)1 (B) dfrac (1)(2) (C)

曲面=dfrac ({x)^2}(2)+(y)^2上与 平面=dfrac ({x)^2}(2)+(y)^2平行的切平面为（ ）=dfrac ({x)^2}

设sim N(mu ,1), X1、X2、X来自总体样本, hat (mu )=dfrac (2)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+

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

(B) dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)-|||-(C) dfrac (1)(2)(X)_(1)+dfrac (1