Question

解方程:(1999-x)^3+(x-1998)^3=1

Answer 1

解: 设x-1998=y
原式=(1-y)^3+y^3=1
(1-y)^2 (1-y)+y^3=1
y^3+y^2-2y+1-y^3+2y^2-y=1
3y^2-3y=0
y^2-y=0
y(y-1)=0
y1=0 y2=1
则①x-1998=0 ②x-1998=1
x=1998 x=1999

Answer 2

(1999-x)^3+(x-1998)^3=1
=[(1999-x)+(x-1998)][(1999-x)^2-(1999-x)(x-1998)+(x-1998)^2]=1
(1999-x)^2-(1999-x)(x-1998)+(x-1998)^2-1=0
(1999-x)^2-(1999-x)(x-1998)+(x-1998-1)(x-1998+1)=0
(1999-x)^2-(1999-x)(x-1998)+(x-1999)(x-1997)=0
(1999-x)^2-(1999-x)(x-1998)-(1999-x)(x-1997)=0
(1999-x)[(1999-x)-(x-1998)-(x-1997)]=0
(1999-x)(1999-x-x+1998-x+1997)=0
(1999-x)(5994-3x)=0
x1=1999
x2=5994/3=1998