Giải thích các bước giải:
a,
\[{5^{2005}} + {5^{2003}} = {5^{2003}}.\left( {{5^2} + 1} \right) = {26.5^{2003}} \vdots 13\]
b,
\(\begin{array}{l}
{\left( {a - b} \right)^2} \ge 0 \Leftrightarrow {a^2} - 2ab + {b^2} \ge 0 \Leftrightarrow {a^2} + {b^2} \ge 2ab\\
{\left( {a - 1} \right)^2} \ge 0 \Leftrightarrow {a^2} - 2a + 1 \ge 0 \Leftrightarrow {a^2} + 1 \ge 2a\\
{\left( {b - 1} \right)^2} \ge 0 \Leftrightarrow {b^2} - 2b + 1 \ge 0 \Leftrightarrow {b^2} + 1 \ge 2b\\
\Rightarrow \left( {{a^2} + {b^2}} \right) + \left( {{a^2} + 1} \right) + \left( {{b^2} + 1} \right) \ge 2ab + 2a + 2b\\
\Leftrightarrow {a^2} + {b^2} + 1 \ge ab + a + b
\end{array}\)
c,
\(\begin{array}{l}
{a^3} + {b^3} + {c^3} - 3abc\\
= {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) + {c^3} - 3abc\\
= \left[ {{{\left( {a + b + c} \right)}^3} - 3\left( {a + b} \right)c\left( {a + b + c} \right)} \right] - 3ab\left( {a + b + c} \right)\\
= {\left( {a + b + c} \right)^3} - \left( {a + b + c} \right)\left( {3ac + 3ab} \right) - 3ab\left( {a + b + c} \right)\\
= {\left( {a + b + c} \right)^3} - \left( {a + b + c} \right)\left( {3ab + 3bc + 3ca} \right)\\
= 0\,\,\,\,\left( {a + b + c = 0} \right)\\
\Rightarrow {a^3} + {b^3} + {c^3} = 3abc
\end{array}\)
