Ta có: \(a+b=\dfrac{-1+\sqrt{2}}{2}+\dfrac{-1-\sqrt{2}}{2}=-1\)
\(ab=\dfrac{-1+\sqrt{2}}{2}.\dfrac{-1-\sqrt{2}}{2}=\dfrac{-1}{4}\)
\(\left(a+b\right)^2=a^2+b^2+2ab=1\)
\(\Rightarrow a^2+b^2+2.\dfrac{-1}{4}=1\)\(\Rightarrow a^2+b^2=\dfrac{3}{2}\) (1)
\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=-1\)
\(\Rightarrow a^3+b^3+3.\dfrac{-1}{4}.\left(-1\right)=-1\)\(\Rightarrow a^3+b^3=\dfrac{-7}{4}\) (2)
Từ (1) và (2) suy ra \(\left(a^2+b^2\right)\left(a^3+b^3\right)=\dfrac{3}{2}.\dfrac{-7}{4}=\dfrac{-21}{8}\)
\(\Rightarrow a^5+a^3b^2+a^2b^3+b^5=\dfrac{-21}{8}\)
\(\Rightarrow a^5+b^5+a^2b^2\left(a+b\right)=\dfrac{-21}{8}\)
\(\Rightarrow a^5+b^5+\dfrac{1}{16}.\left(-1\right)=\dfrac{-21}{8}\)\(\Rightarrow a^5+b^5=\dfrac{-41}{16}\) (3)
Từ (1) và (3) suy ra \(\left(a^2+b^2\right)\left(a^5+b^5\right)=\dfrac{3}{2}.\dfrac{-41}{16}=\dfrac{-123}{32}\)
\(\Rightarrow a^7+a^5b^2+a^2b^5+b^7=\dfrac{-123}{32}\)
\(\Rightarrow a^7+b^7+a^2b^2\left(a^3+b^3\right)=\dfrac{-123}{32}\)
\(\Rightarrow a^7+b^7+\dfrac{1}{16}.\dfrac{-7}{4}=\dfrac{-123}{32}\)\(\Rightarrow a^7+b^7=\dfrac{-239}{64}\)
Vậy \(a^7+b^7=\dfrac{-239}{64}\)