Cho hai số thực khác nhau a, b thỏa mãn: $\frac{1}{a^2+1}+\frac{1}{b^2+1}=\frac{2}{1+ab}$

Tính giá trị của biểu thức: M = $\frac{1}{a^{2021}+1}+\frac{1}{b^{2021}+1}$

Theo giả thiết, ta có: $\frac{1}{a^2+1}+\frac{1}{b^2+1}=\frac{2}{1+ab}$

$$\begin{gathered} \iff \frac{1}{a^2+1}+\frac{1}{b^2+1}-\frac{2}{1+ab}=0 \\ \iff \frac{1}{a^2+1}-\frac{1}{ab+1}+\frac{1}{b^2+1}-\frac{1}{ab+1}=0 \\ \iff \frac{1.\left(ab+1\right)}{\left(a^2+1\right)\left(ab+1\right)}-\frac{1.\left(a^2+1\right)}{\left(ab+1\right)\left(a^2+1\right)}+\frac{1.\left(ab+1\right)}{\left(b^2+1\right)\left(ab+1\right)}-\frac{1.\left(b^2+1\right)}{\left(ab+1\right)\left(b^2+1\right)}=0 \end{gathered}$$

$$\begin{gathered} \iff \frac{\left(ab+1\right)-\left(a^2+1\right)}{\left(a^2+1\right)\left(ab+1\right)}+\frac{\left(ab+1\right)-\left(b^2+1\right)}{\left(b^2+1\right)\left(ab+1\right)}=0 \\ \iff \frac{ab-a^2}{\left(a^2+1\right)\left(ab+1\right)}+\frac{ab-b^2}{\left(b^2+1\right)\left(ab+1\right)}=0 \\ \iff \frac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}-\frac{b\left(b-a\right)}{\left(b^2+1\right)\left(ab+1\right)}=0 \\ \iff \frac{a\left(b-a\right)\left(b^2+1\right)-b\left(b-a\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}=0 \end{gathered}$$

$\iff$ (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 = $\frac{1}{b}$

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

$$\begin{gathered} =\frac{1}{\frac{1}{b^{2021}}+1}+\frac{1}{b^{2021}+1} \\ =\frac{1}{\frac{1}{b^{2021}}+\frac{b^{2021}}{b^{2021}}}+\frac{1}{b^{2021}+1} \\ =\frac{1}{\frac{1+b^{2021}}{b^{2021}}}+\frac{1}{b^{2021}+1} \\ =\frac{b^{2021}}{b^{2021}+1}+\frac{1}{b^{2021}+1} \\ =\frac{b^{2021}+1}{b^{2021}+1}=1 \end{gathered}$$