角妒Y;整合^构铑yb译

AB中点C(x,y)xA+xB=2x,yA+yB=2yk(AB)=(yA-yB)/(xA-xB)=(y-2)/(x-3)[(xA)^2+(yA)^2]-[(xB)^2+(yB)^2]=4-4=0(xA+xB)*(xA-xB)+(yA+yB)*(yA-yB)=02x+2y*(yA-yB)/(xA-xB)=0x+y*(y-2)/(x-3)=0ab中点的轨迹方程是园:(x-1.5)^2+(y-1)^2=2.25

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原式=(x+y)x2(x+y)(xy)+y2xy-2xy(xy)(xy)2=x2xy+y2xy-2xyxy=x2+y22xyxy=x-y.故选B.

其中Y:Ba:Cu=1:2:3所以应该取Y2O3:BaCO3:Cu0=(1/2):(2/2):(3/1)=1:4:6

(仅供参考)过A作AD//x,交准线于D,因为AD=Af,CB=BF,用平面几何证出角CDF=90.勾股定理得出:CD^2=DF^2+CF^2,得到yA*yB=-p^2.要证:OA斜率=OC斜率,即证:yA/xA=yC/xC=yB/(-p/2),只需证(-p/2)yA=xAyB.两边乘以yA:(-p/2)yAyA=xAyByA,因上述的yA*yB=-p^2.只需证(-p/2)yAyA==-p^2xAyA^2=2pxA

选D事件X=Y 有两种可能X=Y =0或X=Y =1他们的概率分别是0.3*0.3=0.9和0.7*0.7=0.49所以P{X=Y}=0.9+0.49=0.58

高脚杯状的.横截面是圆,纵截面是抛物线

( 1 x - 1 y )(xy)= 1 x xy- 1 y xy=y-x,故选C.

先利用函数乘积的求导法则和积分上限函数的求导法则求出导函数,y′=2xex2∫x0et2dx+1,代入得2xy+1=y′,即y′-2xy=1.故选:B.