Đáp án:
\[K = 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{a^{2018}} + {b^{2018}} = {a^{2019}} + {b^{2019}}\\
\Leftrightarrow {a^{2018}} - {a^{2018}} = {b^{2019}} - {b^{2018}}\\
\Leftrightarrow {a^{2018}}\left( {1 - a} \right) = {b^{2018}}\left( {b - 1} \right)\\
\Leftrightarrow \frac{{1 - a}}{{b - 1}} = \frac{{{b^{2018}}}}{{{a^{2018}}}}\\
{a^{2019}} + {b^{2019}} = {a^{2020}} + {b^{2020}}\\
\Leftrightarrow {a^{2019}} - {a^{2020}} = {b^{2020}} - {b^{2019}}\\
\Leftrightarrow {a^{2019}}\left( {1 - a} \right) = {b^{2019}}\left( {b - 1} \right)\\
\Leftrightarrow \frac{{1 - a}}{{b - 1}} = \frac{{{b^{2019}}}}{{{a^{2019}}}}\\
\Rightarrow \frac{{{b^{2018}}}}{{{a^{2018}}}} = \frac{{{b^{2019}}}}{{{a^{2019}}}} \Leftrightarrow \frac{b}{a} = 1 \Leftrightarrow a = b\\
\Rightarrow 2{a^{2018}} = 2{a^{2019}} \Rightarrow a = b = 1\\
\Rightarrow K = {a^{2021}} + {b^{2021}} = 1 + 1 = 2
\end{array}\)
