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

$\iff$ (b – a). (ab² + a) − (b − a). (a²b + b) = 0

$\iff$ (b – a). (ab² − a²b + a − b) = 0

$\iff$ (b – a). [ab. (b – a) – (b – a)] = 0

$\iff$ (b – a). (b – a). (ab – 1) = 0

Vì a ≠ b nên b – a ≠ 0

Do đó (b – a). (b – a). (ab – 1) = 0

$\iff$ ab – 1 = 0

$\iff$ ab = 1

$\iff$ a = $\frac{1}{b}$

$$\begin{gathered} M=\frac{1}{a^{2021}+1}+\frac{1}{b^{2021}+1} \\ =\frac{1}{{\left(\frac{1}{b}\right)}^{2021}+1}+\frac{1}{b^{2021}+1} \end{gathered}$$