Question

[Cô Độc]

Cho a,b dương và \(a^{2020}\)+ \(b^{2020}\)= \(a^{2021}\) + \(b^{2021}\) = \(a^{2022}\) + \(b^{2022}\) . Tính \(a^{2020}\) + \(b^{2021}\)
 

Answer 1

[☆MĭηɦღAηɦ❄]

\(a^{2020}+b^{2020}=a^{2021}+b^{2021}=a^{2022}+b^{2022}\)       (1)

Ta có : \(a^{2021}+b^{2021}=a^{2022}+b^{2022}\)

\(\Leftrightarrow a^{2021}+b^{2021}=a^{2022}+a^{2021}b+b^{2022}+ab^{2021}-a^{2021}b-ab^{2021}\)

\(\Leftrightarrow a^{2021}+b^{2021}=a^{2021}\left(a+b\right)+b^{2021}\left(a+b\right)-ab\left(a^{2020}+b^{2020}\right)\)

\(\Leftrightarrow a^{2021}+b^{2021}=\left(a^{2021}+b^{2021}\right)\left(a+b\right)-ab\left(a^{2020}+b^{2020}\right)\)

\(\Leftrightarrow a+b-ab=1\)

\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-1=0\\1-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}}\)

(+) Thay \(a=1\)vào \(\left(1\right)\)ta được : 

\(b^{2020}=b^{2021}=b^{2022}\Leftrightarrow\orbr{\begin{cases}b=0\\b=1\end{cases}\Leftrightarrow}b=1\left(b>0\right)\)

(+) Thay \(b=1\)vào (1) ta được : 

\(a^{2020}=a^{2021}=a^{2022}\Leftrightarrow\orbr{\begin{cases}a=1\\a=0\end{cases}\Leftrightarrow}a=1\left(a>0\right)\)

\(\Rightarrow a=b=1\)\(\Rightarrow a^{2020}+b^{2021}=1^{2020}+1^{2021}=2\)