Question

Biết \(a^2+b^2=1;a^4+b^4=\frac{1}{2}.\). Tính \(a^{2020}+b^{2020}\)

Answer 1

Ta có: 

\(a^2+b^2=1\Leftrightarrow\left(a^2+b^2\right)^2=1\Leftrightarrow a^4+b^4+2a^2b^2=1\)

\(\Leftrightarrow a^2b^2=\frac{1}{4}\Leftrightarrow b^2=\frac{1}{4a^2}\)

=> \(a^2+\frac{1}{4a^2}=1\Leftrightarrow4a^4-4a^2+1=0\Leftrightarrow\left(2a^2-1\right)^2=0\Leftrightarrow a^2=\frac{1}{2}\)

=> \(b^2=\frac{1}{2}\)

=> \(a^{2020}+b^{2020}=\left(a^2\right)^{1010}+\left(b^2\right)^{1010}=\left(\frac{1}{2}\right)^{1010}+\left(\frac{1}{2}\right)^{1010}=2.\frac{1}{2^{1010}}=\frac{1}{2^{2009}}\)