Ta có:
\(P=2a^{2n+1}-3a^{2n}+5a^{2n+1}-7a^{2n}+3a^{2n+1}\)
\(P=\left(2a^{2n+1}+5a^{2n+1}+3a^{2n+1}\right)+\left(-3a^{2n}-7a^{2n}\right)\)
Suy ra: \(P=10a^{2n+1}+\left(-10a\right)^{2n}\)
Mà \(2n⋮2\)còn \(2n+1\)ko chia hết cho 2
Do đó: \(a>0\)thì P>0
Nhầm cái chỗ suy ra:
\(P=10a^{2n+1}+\left(-10\right)a^{2n}\)