Cho hai số thực khác nhau a, b thỏa mãn: 1/a^2+1 + 1/b^2+1 = 2/1+ab

Theo giả thiết, ta có: 1a2+1+1b2+1=21+ab⇔1a2+1+1b2+1-21+ab=0⇔1a2+1-1ab+1+1b2+1-1ab+1=0⇔1.ab+1a2+1ab+1-1.a2+1ab+1a2+1+1.ab+1b2+1ab+1-1.b2+1ab+1b2+1=0

⇔ab+1−a2+1a2+1ab+1+ab+1−b2+1b2+1ab+1=0⇔ab−a2a2+1ab+1+ab−b2b2+1ab+1=0⇔ab−aa2+1ab+1-bb−ab2+1ab+1=0⇔ab−ab2+1−bb−aa2+1a2+1b2+1ab+1=0

⇔ (b – a). (ab² + a) − (b − a). (a²b + b) = 0

⇔ (b – a). (ab² − a²b + a − b) = 0

⇔ (b – a). [ab. (b – a) – (b – a)] = 0

⇔ (b – a). (b – a). (ab – 1) = 0

Vì a ≠ b nên b – a ≠ 0

Do đó (b – a). (b – a). (ab – 1) = 0

⇔ ab – 1 = 0

⇔ ab = 1

⇔ a = 1b

M=1a2021+1+1b2021+1=11b2021+1+1b2021+1

=11b2021+1+1b2021+1=11b2021+b2021b2021+1b2021+1=11+b2021b2021+1b2021+1=b2021b2021+1+1b2021+1=b2021+1b2021+1=1