第一题,用二项式定理展开,得:A=3^5=243,C=(3^3)*5!/(2!3!)=270,E=3*5!/(4!1!)=15,所以,A+C+E=528
第二题:X=2A-B-C (1)
Y=2B-C-A (2)
Z=2C-A-B (3)
(1)-(3),3(C-A)=Z-X
(1)-(2),3(B-A)=Y-X
(3)-(2),3(B-C)=Y-Z
由(B-C)X+(C-A)Y+(A-B)Z
=B(X-Z)+A(Z-Y)+C(Y-X)
=B(1/3)(A-C)+A(1/3)(C-B)+C(1/3)(B-A)
=0
第三题:
(1) (X-3)^2+|Y+1|+Z^2=0
所以,X=3,Y=-1,Z=0
所以,X^2+Y^2+Z^2-XY-YZ-ZX=13
(2) (1/2)[(X-Y)^2+(Y-Z)^2+(Z-X)^2]=13
(3) 上题(1)(2)相等,可以推得X^2+Y^2+Z^2-XY-YZ-ZX=(1/2)[(X-Y)^2+(Y-Z)^2+(Z-X)^2]