bạn tham khảo nha 

a) ta có: 5^2005 + 5^2003 = 5^2003.5^2 + 5^2003
= 5^2003.(25 + 1)
= 26.5^2003
vì 26 chia hết cho 13
suy ra 26.5^2003 chia hết cho 13
hay 5^2005 + 5^2003 chia hết cho 13

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,

\(\begin{array}{l}
{a^3} + {b^3} + {c^3} - 3abc\\
 = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) + {c^3} - 3abc\\
 = \left[ {{{\left( {a + b + c} \right)}^3} - 3\left( {a + b} \right)c\left( {a + b + c} \right)} \right] - 3ab\left( {a + b + c} \right)\\
 = {\left( {a + b + c} \right)^3} - \left( {a + b + c} \right)\left( {3ac + 3ab} \right) - 3ab\left( {a + b + c} \right)\\
 = {\left( {a + b + c} \right)^3} - \left( {a + b + c} \right)\left( {3ab + 3bc + 3ca} \right)\\
 = 0\,\,\,\,\left( {a + b + c = 0} \right)\\
 \Rightarrow {a^3} + {b^3} + {c^3} = 3abc
\end{array}\)