Ta có : ab=1=>a2b2=1
Ta có: \(\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)
=>\(a^5+a^3b^2+a^2b^3+b^5-a-b\)
=>\(a^5+b^5+a+b-a-b\)( do a2b2=1)
=>\(a^5+b^5\)
Vậy \(a^5+b^5=\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)
NHỚ H CHO MÌNH NHÉ!