Question

设X,Y为实数，且满足（X-1)^3+1998(x-1)=-1;(Y-1)^3+1998(Y-1)=1

求X+Y的值

Answer 1

(x-1)^3+1998(x-1)=-1
(y-1)^3+1998(y-1)=1
相加得(x-1)^3+1998(x-1)+(y-1)^3+1998(y-1)=0
[(x-1)^3+(y-1)^3]+[1998(x-1)+1998(y-1)]=0
(x-1+y-1)[(x-1)^2-(x-1)(y-1)+(y-1)^2]+1998*(x-1+y-1)=0
(x+y-2)(x^2+y^2-xy-x-y+1999)=0
故x+y=2或x^2+y^2-xy-x-y+1999=0

Answer 2

（X-1)^3+1998(x-1)=-1;
(Y-1)^3+1998(Y-1)=1
上面两式相加得到：（X-1)^3+ (Y-1)^3+1998(X+Y-2)=0
三次展开(X+Y-2)[（X-1)^2-(X-1)(Y-1)+(Y-1)^2]+1998(X+Y-2)=0
(X+Y-2)[)[（X-1)^2-(X-1)(Y-1)+(Y-1)^2+1998 ]=0
明显看出中括号[]里面的恒大于0，则X+Y-2=0

X+Y=2

追问

过程咩

追答

过程在上面写了哦