Tìm 3 số x, y, z thỏa mãn điều kiện: 2018x – y2 = 2018y – z2 = 2018z – x2 = 2017.

Ta có: 2018x−y2=2018y−z22018y−z2=2018z−x22018z−x2=2018x−y2⇔2018x−y=y−zy+z   12018y−z=z−xz+x   22018z−x=x−yx+y   32018x−y2=2018y−z22018y−z2=2018z−x22018z−x2=2018x−y2⇔2018x−y=y−zy+z   12018y−z=z−xz+x   22018z−x=x−yx+y   3

Lấy (1) . (2) . (3) ta được:

2018³ . (x – y)(y – z)(z – x) = (x – y)(y – z)(z – x)(x + y)(y + z)(z + x)

⇔ 2018³ = (x + y)(y + z)(z + x)

⇔ x + y = y + z = z + x = 2018

⇔ x = y = z

Theo đề bài, ta có:

2018x – y² = 2018y – z² = 2018z – x² = 2017

⇔ 2018x – x² = 2018y – y² = 2018z – z² = 2017

Có: 2018x – x² = 2017

⇔ x² – 2018x + 2017 = 0

⇔x² – x – 2017x + 2017 = 0

⇔x(x – 1) – 2017(x – 1) = 0

⇔ (x – 2017)(x – 1) = 0

⇔x = 2017 hoặc x = 1

Vậy x = y = z = 2017 hoặc x = y = z = 1.