) ( )-|||-(A) =pm (2k+1) (B) =pm 2k-|||-(C) =pm dfrac (1)(2)(2k+1) (D) =pm dfrac (2k+1)(4)

. . .-|||-.-dfrac (2)(3)pi +2kpi ,k=0.pm 1 . . . .-|||-(C) dfrac (1)(3)pi +2kpi

若_(1)+((k)^2+1)(x)_(2)+2(x)_(3)=0-|||-_(1)+(2k+1)(x)_(2)+2(x)_(3)=0-|||-(x)_(1)+

X=k =dfrac (c)(k!) ,=0, 1,2,3...... X=k =dfrac (c)(k!) ,=0, 1,2,3...... X=k =dfr

).，求：（1）参数a； X=k =dfrac (a)({2)^k}(k=1,2,... ).； X=k =dfrac (a)({2)^k}(k=1,2,...

)-|||-B .(X=k)=dfrac (1)({3)^k}(k=1,2,... )-|||-C .(X=k)=dfrac (1)({3)^k}(k=0,1

若随机变量X的分布列为 (X=k)=dfrac (c)(k(k+1)) ,k=1,-|||-2,3,4,其中c为常数,则 (dfrac (1)(2)lt Xlt

设PX=K=dfrac(1)(k(k+1))，k=1，2，···，则E(X)=(,,,,,)A. 0B. 1C. 0.5D. 不存在

M~m-M-|||-k1 k2-|||-(B) =(x)_(0)cos [ sqrt (dfrac {{k)_(1)(k)_(2)}(m({k)_(1)+(k)

证明以下结构晶面族的面间距:(1)立方晶系:_(h{K)_(k)}=a[ (h)^2+(k)^2+(1)^2] -dfrac (1)(2),,(2)正交晶系:_

12.已知公式 =dfrac (pi )(4)(d)^2k 中,d和k为直接测得量,并且 =(5.00pm 0.01)cm , =(10.00pm -|||-0