1 2 Y~-|||-0 1 7-|||-设X~ 1/3 2/3 J 1/2 1/2 X+Y=2 =1/6, 则-|||-(A) P(X=Y )=1/3 (B) X=Y =1/6-|||-(C) X=1+Y =2/3 (D) X=Y+1 =5/6-|||-

6.设 (x,y)=dfrac (x-{y)^2+(y)^3}(2x+{y)^2}, 则,lim h(x,y)等于 ()-|||-(A) dfrac (1)(2

设平面区域D由 x=0 ,y=0 +y=dfrac (1)(2) ,x+y=1 围成,记 _(1)=iint (ln )^3(x+y)dxdy --|||-_(

3.已知两个线性变换-|||- ) (x)_(1)=2(y)_(1)+(y)_(3), (x)_(2)=-2(y)_(1)+3(y)_(2)+2(y)_(3)

曲线y=(x-1 )3√x^2的凹区间为（ ）y=(x-1 )3√x^2y=(x-1 )3√x^2y=(x-1 )3√x^2y=(x-1 )3√x^2曲线的凹区

10.已知线性变换-|||- ) (x)_(1)=2(y)_(1)+2(y)_(2)+(y)_(3) (x)_(2)=3(y)_(1)+(y)_(2)+5(y

3.已知两个线性变换-|||- ) (x)_(1)=2(y)_(1)+ (y)_(3) (x)_(2)=-2(y)_(1)+3(y)_(2)+2(y)_(3

设复数_(1)=(x)_(1)+i(y)_(1), _(2)=(x)_(2)+i(y)_(2)，且_(1)=(x)_(1)+i(y)_(1), _(2)=(x)

求函数y=x3-3x2-9x+1的极值.正确答案:由于y=x3-3x2-9x+1的定义域为(-∞,+∞),y’=3x2-6x-9,令y’=0,得驻点x1=-1,

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

已知两个线性变换： ) (y)_(1)=2(x)_(1)+(x)_(2) (y)_(2)=3(x)_(1)-(x)_(2)+2(x)_(3) (y)_(3)=