THAM KHẢO!                                                                                                                           555²²² + 222⁵⁵⁵ =222⁵⁵⁵ + 555⁵⁵⁵ - (555⁵⁵⁵ - 555²²²) 
= 222⁵⁵⁵ + 555⁵⁵⁵ - 555²²²(555³³³ - 1) 
Ta có :
222⁵⁵⁵ + 555⁵⁵⁵ chia hết cho 222 + 555 = 777 chia hết cho 7 (1) 
555³³³ - 1 = (555³)¹¹¹ - 1 ⋮⋮ 555³ - 1 
Ta có 555 = 7 . 79 + 2 = 7k + 2 (với k = 79) 
555³ - 1 = (7k+2)³ - 1 = (7k)³ + 3.(7k)².2 + 3.7k.2² + 8 - 1 = (7k)³ + 3.(7k)².2 + 3.7k.2² + 7 ⋮⋮ 7 
=> 555³³³ - 1 chia hết cho 7 (2) 
Từ (1) và (2) => 555²²² + 222⁵⁵⁵ chia hết cho 7 (đpcm)

a,333^444 < 444^333

B,5^340 > 7^255

\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{184}+\frac{1}{238}+\frac{1}{340}=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}>\frac{2}{20}=\frac{1}{10}=0,1\)

vậy S>0,1

a) A = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20

=> 3A = 1/2 - 1/5 + 1/5 - .... + 1/14 - 1/17 + 1/17 - 1/20

=> 3A = 1/2 - 1/20 = 9/20

=> A = 3/20

b) 2004¹⁰ + 2004⁹ = 2004⁹(1+2004) = 2004⁹ . 2005

2005¹⁰  = 2005⁹  . 2005

Do 2005⁹ > 2004⁹ nên 2004¹⁰ + 2004⁹ < 2005¹⁰

S = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

S = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)

S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}\)

S = \(\frac{3}{20}\)

S = 0,15 > 0,1

S=\(\frac{3}{20}\)>0,1

li ke cho mình nha

Bài làm

Ta có: 5³⁴⁰ = 5^4.85 = (5⁴)⁸⁵ = 625⁸⁵  

          7²⁵⁵ = 7^3.85 = (7³)⁸⁵ = 343⁸⁵  

Mà 625 > 343 

=> 625⁸⁵ > 343⁸⁵  

Hay 5³⁴⁰ > 7²⁵⁵ 

Vậy 5³⁴⁰ > 7²⁵⁵ 

# Chúc bạn học tốt #

thank

Ta có : \(5^{340}=(5^4)^{85}=625^{85}\)

            \(7^{255}=\left(7^3\right)^{85}=343^{85}\)

            Mà \(625>343\)

\(\Rightarrow625^{85}>343^{85}\)

Vậy \(5^{340}>7^{255}\)

\(1.\)

a, \(27^{265}\)và \(81^{199}\)

\(27^{265}=\left(3^3\right)^{265}=3^{795}\)

\(81^{199}=\left(3^4\right)^{199}=3^{796}\)

\(\Rightarrow3^{795}< 3^{796}hay27^{265}< 81^{199}\)

b, \(1024^{15}=\left(2^{10}\right)^{15}=2^{150}\)

\(128^{21}=\left(2^7\right)^{21}=2^{147}\)

\(2^{150}>2^{147}.hay.1024^{15}>128^{21}\)

\(A=\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+\frac{153}{154}+\frac{237}{238}+\frac{339}{340}=\frac{117}{20}\)

\(suyra:A<1\)

Phân tử khối của 5 phân tử Magie cacbonat MgCO3 là bao nhiêu ?

A. 430

B. 440

C. 420

D. 410

Ta có : \(5M_{MgCO3}=5\left(24+12+16.3\right)=420\)

=> Đáp án C .