Vì tui dùng app giải

\(1;a,942^{60}-351^{37}\)

\(=\left(942^4\right)^{15}-\left(....1\right)\)

\(=\left(....6\right)^{15}-\left(...1\right)\)

\(=\left(...6\right)-\left(...1\right)=\left(....5\right)⋮5\)

\(b,99^5-98^4+97^3-96^2\)

\(=\left(...9\right)-\left(...6\right)+\left(...3\right)-\left(...6\right)\)

\(=\left(...6\right)-\left(...6\right)=\left(...0\right)⋮2;5\)

\(2;5n-n=4n⋮4\)

chả hiểu j

a) \(3^5+3^4+3^3\)

\(=3^3\cdot3^2+3^3\cdot3+3^3\cdot1\)

\(=3^3\left(3^2+3+1\right)\)

\(=3^3\cdot13⋮13\)     (đpcm)

b) \(2^{10}-2^9+2^8-2^7\)

\(=2^7\cdot2^3-2^7\cdot2^2+2^7\cdot2-2^7\cdot1\)

\(=2^7\left(2^3-2^2+2-1\right)\)

\(=2^7\cdot5⋮5\)    (đpcm)

=))

\(A=3^1+3^2+...+3^{30}\)

=> A=3(1+3) +...+ 3²⁹(1+3)

        =3.4+ ... + 3²⁹.4 \(⋮\)4

Chia het 13 ban lam tuong tu nhe

\(2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{99}.3\)

\(=3\left(2+2^3+...+2^{99}\right)\)chia hết cho 3 (Đpcm)

Đặt A = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰

Ta có:
A = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰

A = (2 + 2²) + (2³ + 2⁴) + ... + (2⁹⁹ + 2¹⁰⁰)

A = 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁹⁹.(1 + 2)

A = 2.3 + 2³.3 + ... + 2⁹⁹.3

A = (2 + 2³ + ... + 2⁹⁹) . 3

Vì (2 + 2³ + ... + 2⁹⁹) . 3 chia hết cho 3 nên 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰ chia hết cho 3 (đpcm)

\(2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=\left(2+2^2+2^3+2^4\right)+2^4\cdot\left(2+2^2+2^3+2^4\right)+....+2^{96}\cdot\left(2+2^2+2^3+2^4\right)\)

\(\)\(=30+2^4\cdot30+...+2^{92}\cdot30\)

\(=30\cdot\left(2^4+2^8+...+2^{96}\right)\)

Vì 30 chia hết cho 3 \(\Rightarrow\)

\(2+2^2+2^3+2^4+...+2^{99}+2^{100}\)chia hết cho 3

7^100-7^99+7^98

=7^98(7^2-7+1)

=7^98.43 chia hết cho 43

b) ta có 2^62=(2^2)^31=4^31

vì 4^31<5^31=>2^62<5^31