CMR: \(5^{100}-5^{99}+5^{98}\)chia hết cho 7
Ta có: \(5^{100}-5^{99}+5^{98}\)
\(=5^{98}.5^2-5^{98}.5+5^{98}\)
\(=5^{98}.\left(5^2-5-1\right)\)
\(=5^{98}.21\)
\(=5^{98}.3.7\)
=> \(5^{100}-5^{99}+5^{98}\)chia hết cho 7
\(5^{100}-5^{99}+5^{98}\)
\(=5^{98}.\left(5^2-5+1\right)\)
\(=5^{98}.21\)
\(=5^{98}.3.7\)chia hết cho 7
Theo đầu bài ta có:
\(5^{100}-5^{99}+5^{98}\)
\(\Leftrightarrow5^{98+2}-5^{98+1}+5^{98+0}\)
\(\Leftrightarrow5^{98}\cdot5^2-5^{98}\cdot5^1+5^{98}\cdot5^0\)
\(\Leftrightarrow5^{98}\cdot\left(5^2-5^1+5^0\right)\)
\(\Leftrightarrow5^{98}\cdot\left(25-5+1\right)\)
\(\Leftrightarrow5^{98}\cdot21\)
Do 21 chia hết cho 7 nên 5^98 * 21 chia hết cho 7 => 5^100 - 5^99 + 5^98 chia hết cho 7
