a) sửa đề:CM chia 3 dư 1

chia 103 số hạng ra thành nhóm có 3 số

=>103:3=34( nhóm) và dư 1 số hạng

<=> A=\(1+\left(-2+2^2\right.-2^3)+\left(2^4-2^5+2^6\right)-.\ldots+\left(2^{100}-2^{101}+2^{102}\right)\)

A=\(1-2\left(1-2+2^2\right)+2^4\left(1-2+2^2\right)-.\ldots+2^{100}\left(1-2+2^2\right)\)

\(A=1-2\cdot3+2^4\cdot3-.\ldots+2^{100}\cdot3\)

\(A=1+3\left(-2+2^4-.\ldots+2^{100}\right)\)

vì trong ngoặc chia hết cho 3

=> 1 là số còn lại

vậy A chia 3 dư 1

b) \(A=1-2^1+2^2-2^3+\cdots+2^{100}-2^{101}+2^{102}\)

\(2A=2-2^2+2^3-2^4+\cdots+2^{101}-2^{102}+2^{103}\)

=> \(A+2A=\left(1-2^1+2^2-2^3+\cdots+2^{100}-2^{101}+2^{102}\right)+\left(2-2^2+2^3-2^4+\cdots+2^{103}\right)\)

\(3A=1+2^{103}\)

\(A=\frac{\left(2^{103}+1\right)}{3}\)

\(3,1+5^2+5^4+...+5^{26}\)

\(=\left(1+5^2\right)+\left(5^4+5^6\right)+...+\left(5^{24}+5^{26}\right)\)

\(=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{24}\left(1+5^2\right)\)

\(=26+5^4.26+...+5^{24}.26\)

\(=26\left(5^4+...+5^{24}\right)\)

Vì  \(26⋮26\)

\(\Rightarrow26\left(5^4+...+5^{24}\right)⋮26\)

\(\Rightarrow1+5^2+5^4+...+5^{26}⋮26\)

\(4,1+2^2+2^4+...+2^{100}\)

\(=\left(1+2^2+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)

\(=\left(1+2^2+2^4\right)+....+2^{98}\left(1+2^2+2^4\right)\)

\(=21+2^6.21...+2^{98}.21\)

\(=21\left(2^6+...+2^{98}\right)\)

Có : \(21\left(2^6+...+2^{98}\right)⋮21\)

\(\Rightarrow1+2^2+2^4+...+2^{100}⋮21\)

a) 5^x=5^78:5^14(lấy 78-14)

5^x=5^64

=> x=64

b) 7^x.7^2=7^21

7^x=7^21:7^2

7^x=7^19

=> x=19

1. a) có 5^x . 5¹⁴ = 5⁷⁸

Hay: 5^x+14 =5⁷⁸

^{=> x+14=78}

=> x=64