b) Ta có: \(B=\dfrac{7}{8}\cdot\dfrac{4}{9}+\dfrac{7}{8}\cdot\dfrac{5}{9}\)

\(=\dfrac{7}{8}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{7}{8}\cdot1=\dfrac{7}{8}\)

\(A=5^5+5^4-8.5^3=5^3\left(5^2+5-8\right)\)

\(=5^3.22=5^3.2.11\)\(⋮\)\(11\)

Vậy  A chia hết cho 11

Ta có : A = 5⁵ + 5⁴ - 8 .5³

              = 5³.(5² + 5 - 8)

             = 5³. 22 

Vì 22 = 2 . 11 \(⋮\)11 => 5³ . 22 \(⋮\)11

Vậy A \(⋮\)11

Mik không chắc cho lắm :

Theo bài ra ta có : A = 5⁵ + 5⁴ − 8 . 5³ = 5³ .( 5² + 5 − 8 ) = 5³ . 22 = ( 5³ . 2 ) .11

Vì 11 chia hết cho 11 nên A chia hết cho 11 .

\(=5^3\left(25+5-8\right)=5^3.2.11\)

=> dpcm

\(A=5^5+5^4-8.5^3\)

\(A=5^5+5^4-2^3.5^3\)

\(A=5^5+5^4-10^3\)

\(A=3750-1000\)

\(A=2750\)chia hết \(11\) ( vì ( 2 + 5 ) - ( 7 + 0 ) = 0 chia hết 11 )

\(A=5^5+5^4-8.5^3\)

\(A=5^3.\left(5^2+5-8\right)\)

\(A=5^3.22\)

\(A=5^3.2.11⋮11\)

Vay A chia het cho 11

tính ra rồi chia cho 11 thôi=1250

A=5⁵+5⁴-8.5³ \(\Leftrightarrow\)A=5³.(5²+5+8) \(\Leftrightarrow\)A=5³.22

\(\Leftrightarrow\)A=5³.2.11

Vì 11 \(⋮\)11 \(\Rightarrow\)A=5³.2.11 \(⋮\)11 hay A=5⁵+5⁴-8.53  \(⋮\)11  (đpcm)

\(A=5^5+5^4-8.5^3\)

\(A=5^3.\left(5^2+5-8\right)\)

\(A=5^3.22\)

\(A=5^3.2.11⋮11\)

Vậy \(A=5^5+5^4-8.5^3\)chia hết cho 11

a) Đặt \(C=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)

\(\Rightarrow5C=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)

\(\Rightarrow5C-C=1-\dfrac{1}{5^{100}}\Rightarrow4C=1-\dfrac{1}{5^{100}}\Rightarrow C=\dfrac{1-\dfrac{1}{5^{100}}}{4}\)

\(\Rightarrow A=8.5^{100}.\dfrac{1-\dfrac{1}{5^{100}}}{4}+1=2.\left(5^{100}-1\right)+1=2.5^{100}-2+1=2.5^{100}-1\)

b)\(B=\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)

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

Đặt \(\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)=D\)

\(\Rightarrow3D=1-\dfrac{1}{3}+...-\dfrac{1}{3^{99}}\)

\(\Rightarrow3D+D=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow D=\dfrac{1-\dfrac{1}{3^{100}}}{4}\)

a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)

\(=8\cdot5^2+5-1\)

\(=200+4=204\)

b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)

\(=27-1+125-1\)

=150

a. (8,5⁷ - 5⁵ + 5⁶) : 5⁵

= (8,5⁷ : 5⁵) - (5⁵ : 5⁵) + (5⁶ : 5⁵)

= 1,7² - 1 + 5

= 2,89 - 1 + 5

= 6,89

b. (9³⁰ - 27¹⁹) : 3⁵⁷ + (125⁹ - 25¹²) : 5²⁴

= (9³⁰ : 3⁵⁷) - (27¹⁹ : 3⁵⁷) + (125⁹ : 5²⁴) - (25¹² : 5²⁴)

= 3³ - 1 + 125 - 1

= 27 - 1 + 125 - 1

= 150

c. (10¹² + 5¹¹ . 2⁹ - 5¹³ - 2⁸) : 4 . 5⁵ . 10⁶

= (10¹² + 2,5 , 10¹⁰ - 5¹³ - 2⁸) : 1,25 . 10¹⁰

= (10¹² : 1,25 . 10¹⁰) + (2,5 . 10¹⁰ : 1,25 . 10¹⁰) - (5¹³ : 1,25 . 10¹⁰) - (2⁸ : 1,25 . 10¹⁰)

= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)

= 81,90234373 \(\approx\) 82

\(\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^8\cdot5^7\left(2^3+5^3\right)}{2^5\cdot5^4\left(2^3+5^3\right)}\\ =\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\\ =2^3\cdot5^3\\ =8\cdot125\\ =1000\)