解方程。 dfrac (1)(2)x－5＝ dfrac (1)(2) 5y－2y＝18 dfrac (1)(2)＋x＝ dfrac (1)(2)

dfrac (1)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+dfrac (2)(5)(X)_(3)-|||-D .dfrac (1)(7)(

直线dfrac (x-1)(1)=dfrac (y)(2)=dfrac (z+3)(-2)-|||-__ __-|||-__与平面dfrac (x-1)(1)=

(B) dfrac (x)(y)((y+1))^2-|||-(C) ^2((x+dfrac {1)(x))}^2. (D) dfrac (y)(x)((y+1)

直线dfrac (x-1)(2)=dfrac (y)(-1)=dfrac (z-2)(3)-|||-__与直线dfrac (x-1)(2)=dfrac (y)(

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

设L为椭圆dfrac ({x)^2}(2)+dfrac ({y)^2}(3)=1，其周长为a，则dfrac ({x)^2}(2)+dfrac ({y)^2}(3

求椭球面dfrac ({x)^2}(2)+dfrac ({y)^2}(3)+dfrac ({z)^2}(4)=1上点dfrac ({x)^2}(2)+dfrac

(B) dfrac (1)(2)(X)^2+dfrac (1)(2)(Y)^2 服从x^2分布.-|||-(C) dfrac (1)(3)((X+Y))^2 服

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

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