Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = {a^4} - 4{a^3} + 5{a^2} - 4a + 5\\
= \left( {{a^4} - 4{a^3} + 4{a^2}} \right) + \left( {{a^2} - 4a + 4} \right) + 1\\
= {a^2}.\left( {{a^2} - 4a + 4} \right) + \left( {{a^2} - 4a + 4} \right) + 1\\
= \left( {{a^2} - 4a + 4} \right)\left( {{a^2} + 1} \right) + 1\\
= \left( {{a^2} - 2.a.2 + {2^2}} \right)\left( {{a^2} + 1} \right) + 1\\
= {\left( {a - 2} \right)^2}.\left( {{a^2} + 1} \right) + 1\\
{\left( {a - 2} \right)^2} \ge 0,\,\,\,\forall a\\
{a^2} + 1 \ge 1 > 0,\,\,\,\forall a\\
\Rightarrow P = {\left( {a - 2} \right)^2}\left( {{a^2} + 1} \right) + 1 \ge 1 > 0,\,\,\,\forall a
\end{array}\)
