a) \(\left(1+2+2^2+...+2^7\right)\)

\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^6+2^7\right)\)

\(=\left(1+2\right)+2^2.\left(1+2\right)+...+2^6.\left(1+2\right)\)

\(=3+2^2.3+...+2^6.3\)

\(=3.\left(1+2^2+...+2^6\right)⋮3\left(đpcm\right)\)

a) Đặt A = 1 + 2 + 2² + 2³ + ... + 2⁷

Ta có:

A = 1 + 2 + 2² + 2³ + ... + 2⁷

\(\Rightarrow\)2A = 2 + 2² + 2³ + 2⁴ + ... + 2⁸

\(\Rightarrow\)A = 2⁸ - 1 = 255

Vì 255\(⋮\)3\(\Rightarrow\)2 + 2² + 2³ + 2⁴ + ... + 2⁸\(⋮\)3

\(\Rightarrow\)ĐPCM

a: Đặt \(A=2+2^2+\cdots+2^{60}\)

Ta có: \(A=2+2^2+\cdots+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{59}+2^{60}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+\cdots+2^{59}\right)\) ⋮3

Ta có: \(A=2+2^2+\cdots+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+\cdots+2^{58}\right)\) ⋮7

TA có: \(A=2+2^2+\cdots+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2_{}^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(1+2^5+\cdots+2^{57}\right)\) ⋮15

b: Ta có: \(B=1+3+3^2+\cdots+3^{1991}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\cdots+\left(3^{1989}+3^{1990}+3^{1991}\right)\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+\cdots+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+\cdots+3^{1989}\right)\) ⋮13

c: Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{1997}+3^{1998}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+\cdots+3^{1996}\left(3+3^2\right)\)

\(=12\left(1+3^2+\cdots+3^{1996}\right)\) ⋮12

Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{1996}+3^{1997}+3^{1998}\right)\)

\(=\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+\cdots+3^{1995}\left(3+3^2+3^3\right)\)

\(=39\left(1+3^3+\cdots+3^{1995}\right)\) ⋮39

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{3}{4}+\dfrac{2}{3}+\dfrac{3}{5}\)

`=`\(\dfrac{9}{12}+\dfrac{8}{12}+\dfrac{3}{5}\)

`=`\(\dfrac{17}{12}+\dfrac{3}{5}\)

`=`\(\dfrac{85}{60}+\dfrac{36}{60}\)

`=`\(\dfrac{121}{60}\)

`b)`

\(\dfrac{1}{2}\cdot\dfrac{9}{13}\div\dfrac{27}{26}\)

`=`\(\dfrac{1}{2}\cdot\dfrac{9}{13}\cdot\dfrac{26}{27}\)

`=`\(\dfrac{1}{2}\cdot\dfrac{2}{3}\)

`=`\(\dfrac{1}{3}\)

`c)`

\(\dfrac{2}{7}\cdot\dfrac{1}{9}+\dfrac{2}{7}\cdot\dfrac{2}{9}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)

`=`\(\dfrac{2}{7}\cdot\left(\dfrac{1}{9}+\dfrac{2}{9}\right)+\dfrac{1}{3}\cdot\dfrac{5}{7}\)

`=`\(\dfrac{2}{7}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)

`=`\(\dfrac{1}{3}\cdot\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)

`=`\(\dfrac{1}{3}\cdot1=\dfrac{1}{3}\)

`d)`

\(11\div\dfrac{5}{2}+11\div\dfrac{7}{3}+11\div\dfrac{35}{6}\)

`=`\(11\cdot\dfrac{2}{5}+11\cdot\dfrac{3}{7}+11\cdot\dfrac{6}{35}\)

`=`\(11\cdot\left(\dfrac{2}{5}+\dfrac{3}{7}+\dfrac{6}{35}\right)\)

`=`\(11\cdot1=11\)

a) 3/4 + 2/3 + 3/5 = 45/60 + 40/60 + 36/60 = 121/60

b) 1/2 x 9/13 : 27/26 = 9/26 x 26/27 = 1/3

c) 2/7 x 1/9 + 2/7 x 2/9 + 1/3 x 5/7 = 2/7 x (1/9 + 2/9) + 5/21 = 2/7 x 1/3 + 5/21 = 2/21 + 5/21 = 1/3

d) 11 : 5/2 + 11 : 7:3 + 11 : 35/6 = 11 x (2/5 + 3/7 + 6/35) = 11 x 1 = 11 

1) Ta có: \(2⋮n-3\)

\(\Leftrightarrow n-3\inƯ\left(2\right)\)

\(\Leftrightarrow n-3\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{4;2;5;1\right\}\)

Vậy: \(n\in\left\{4;2;5;1\right\}\)

2) Ta có: \(n+2⋮n-3\)

\(\Leftrightarrow n-3+5⋮n-3\)

mà \(n-3⋮n-3\)

nên \(5⋮n-3\)

\(\Leftrightarrow n-3\inƯ\left(5\right)\)

\(\Leftrightarrow n-3\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{4;2;8;-2\right\}\)

Vậy: \(n\in\left\{4;2;8;-2\right\}\)

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