Giải thích các bước giải:
a,
Ta có:
\[\begin{array}{l}
xy - x = 15\\
\Leftrightarrow x\left( {y - 1} \right) = 15\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 1\\
y - 1 = 15
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 15\\
y - 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 3\\
y - 1 = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 5\\
y - 1 = 3
\end{array} \right.
\end{array} \right.
\end{array}\]
b,
Ta có:
\[\begin{array}{l}
A = 1 + 5 + {5^2} + {5^3} + .... + {5^{19}} + {5^{20}}\\
\Leftrightarrow A = \left( {1 + 5 + {5^2}} \right) + \left( {{5^3} + {5^4} + {5^5}} \right) + ..... + \left( {{5^{18}} + {5^{19}} + {5^{20}}} \right)\\
\Leftrightarrow A = \left( {1 + 5 + {5^2}} \right) + {5^3}\left( {1 + 5 + {5^2}} \right) + .... + {5^{18}}\left( {1 + 5 + {5^2}} \right)\\
\Leftrightarrow A = \left( {1 + 5 + {5^2}} \right)\left( {1 + {5^3} + .... + {5^{18}}} \right)\\
\Leftrightarrow A = 31\left( {1 + {5^3} + ... + {5^{18}}} \right) \vdots 31
\end{array}\]
