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Trần An

$a^2+b^2+ab-a+b+1=0$

$2a^2+2b^2+2ab-2a+2b+2=0$

$\left(a^2-2a+1\right)+\left(b^2+2b+1\right)+\left(a^2+2ab+b^2\right)=0$

$\left(a-1\right)^2+\left(b+1\right)^2+\left(a+b\right)^2=0$

Vì $\left(a-1\right)^2\ge0;\left(b+1\right)^2\ge0;\left(a+b\right)^2\ge0$

Để $\left(a-1\right)^2+\left(b+1\right)^2+\left(a+b\right)^2=0$ thì

$\begin{cases}\left(a-1\right)^2=0 \\ \left(b+1\right)^2=0 \\ \left(a+b\right)^2=0\end{cases}\Leftrightarrow\begin{cases}a=1 \\ b=-1\end{cases}$

$N=3a^3-2b^4+2024$

$=3.1^3-2.\left(-1\right)^4+2024$

$=3-2+2024$

$=2025$.