Ta có: a+b+c=2020
\(\Leftrightarrow\left\{{}\begin{matrix}a=2020-b-c\\b=2020-a-c\\c=2020-b-a\end{matrix}\right.\)
Ta có: \(P=\left(ab+c-2019\right)\left(bc+a-2019\right)\left(ca+b-2019\right)\)
\(=\left(ab+2020-a-b-2019\right)\left(bc+2020-b-c-2019\right)\left(ca+2020-a-c-2019\right)\)
\(=\left(ab-a-b+1\right)\left(bc-b-c+1\right)\left(ca-a-c+1\right)\)
\(=\left[a\left(b-1\right)-\left(b-1\right)\right]\left[b\left(c-1\right)-\left(c-1\right)\right]\left[a\left(c-1\right)-\left(c-1\right)\right]\)
\(=\left(b-1\right)\left(a-1\right)\left(c-1\right)\left(b-1\right)\left(c-1\right)\left(a-1\right)\)
\(=\left[\left(a-1\right)\left(b-1\right)\left(c-1\right)\right]^2\)
Vậy: P là số chính phương(đpcm)
