\(S=1+2+2^2+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow S=2^{10}-1\)

Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)

Vậy \(S< 5.2^8\)

S=1+2+2^2+2^3+...+2^9​

2S=2+2^2+2^3+...+2^9+2^10

2S-S=(2+2^2+2^3+...+2^9+2^10)-(1+2+2^2+2^3+...+2^9)

S=2^10-1

5.2^8=(2^2+1).2^8=(2^2.2^8)+(1.2^8)=2^10+2^8

Vì 2^10-1<2^10+2^8=> S<5.2^8

Vậy S < 5. 2^8

Ta có: \(S=1+2+2^2+2^3+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{10}\right)-\left(1+2+2^2+...+2^9\right)\)

\(\Rightarrow S=2^{10}-1\)

Mặt khác: \(5.2^8=\left(1+2^2\right).2^8=2^8+2^2.2^8=2^8+2^{10}\)

Vì \(2^{10}-1< 2^8+2^{10}\Rightarrow S< 5.2^8\)

(2^4 x 5^2 x 11^2 x 7) : (2^3 x 5^3 x 7^2 x 11)

= (2^3 x 2 x 5^2 x 11 x 77) : ( 2^3 x 5^2 x 5 x 7 x 77)          bỏ những số trùng nhau vì là phép nhân

= (2 x 11) : (5 x 7) 

= 22/35

http://123link.pw/j6KCoe

   5⁹⁹ - 4² x 5⁹⁷ - 3² x 5⁹

= 5⁹⁷.(5² - 4²) - 3².5⁹

= 5⁹⁷.(25 - 16) - 3².5⁹

= 5⁹⁷.9 - 9.5⁹

b, 2^y+1 + 2^y+2 = 24

   2^y+1.(1 + 2) = 24

   2^y+1.3 = 24

   2^y+1   = 24: 3

   2^y+1 = 8

   2^y+1 = 2³

  y + 1 = 3

   y = 2 

A=2(1+2+2²+...+2¹²) 

=> A chia hết cho 2

Vậy A chia hết cho 2(đpcm)

a chia hết cho 2

A = 2 + 2² + 2³ + ... + 2¹¹ + 2¹²

   = 2.(1 + 2 + 2² + ... + 2¹⁰ + 2¹¹) chia hết cho 2

Vậy A chia hết cho 2 (ĐPCM).

Giúp mình với mai mình thi rồi

\(ab+ba=(10a+b)+(10b+a)\)

\(=10a+b+10b+a\)

\(=11a+11b\)

\(=11\left(a+b\right)\)

\(a+b\inℕ\Rightarrow ab+ba⋮11\)

\(A=2+2^2+2^3+\cdot\cdot\cdot+2^{2008}\)

\(\Rightarrow2A=2^2+2^3+2^4+\cdot\cdot\cdot+2^{2009}\)
\(\Rightarrow2A-A=\left(2^2+\cdot\cdot\cdot2^{2009}\right)-\left(2+\cdot\cdot\cdot+2^{2008}\right)\)

\(\Rightarrow A=2^{2009}-2\)

a) \(\left(3^4.57-9^2.21\right):3^5\)

\(=\left(3^4.57-3^4.21\right):3^5\)

\(=\left[3^4\left(57-21\right)\right]:3^5\)

\(=3^4.36:3^5\)

\(=3^4.2^2.3^2:3^5\)

\(=3.4\)

\(=12\)

b) Ta có; \(1^3+2^3+...+9^3=2025\)

\(\Leftrightarrow2^3.\left(1^3+2^3+....+9^3\right)=2^3.2025\)

\(\Leftrightarrow2^3+4^5+...+18^3=16200\)