Cho 1/x + 1/y + 1/z = 1/(x+y+z). Chứng minh rằng: 1/x^2023 + 1/y^2023 + 1/z^2023 = 1/(x^2023 + y^2023+z^2023)
Giải thích
Theo giả thiết, 1x+1y+1z=1x+y+z.
Suy ra yz+xz+xyxyz=1x+y+z.
yz+xz+xyx+y+z=xyz
yzx+y+z+xzx+y+z+xyx+y+z=xyz
xyz+y2z+yz2+x2z+xyz+xz2+x2y+xy2+xyz=xyz
x2z+2xyz+y2z+yz2+xz2+x2y+xy2=0
zx+y2+z2x+y+xyx+y=0
x+yzx+y+z2+xy=0
x+yxz+yz+z2+xy=0
x+yxz+xy+yz+z2=0
x+yxy+z+zy+z=0
x+yy+zx+z=0
Suy ra x+y=0 hoặc y+z=0 hoặc x+z=0.
⦁ Nếu x+y=0 thì x=−y, khi đó x2023=−y2023.
Ta có 1x2023+1y2023+1z2023=1−y2023+1y2023+1z2023=1z2023;
1x2023+y2023+z2023=1−y2023+y2023+z2023=1z2023.
Do đó 1x2023+1y2023+1z2023=1x2023+y2023+z2023.
⦁ Nếu y+z=0 hoặc x+z=0, chứng minh tương tự ta cũng có
1x2023+1y2023+1z2023=1x2023+y2023+z2023.
Vậy nếu 1x+1y+1z=1x+y+z thì 1x2023+1y2023+1z2023=1x2023+y2023+z2023.