-9/6

=\(-\dfrac{3}{2}\)

a: =>5x+25-3x+6=25+18

=>2x+41=43

=>2x=2

=>x=1

b: =>4x+8=3x+3+17

=>4x+8=3x+20

=>x=12

a: =>5x+25-3x+6=25+18

=>2x+41=43

=>2x=2

=>x=1

b: =>4x+8=3x+3+17

=>4x+8=3x+20

=>x=12

h: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

m: \(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

\(A=3+3^2+...+3^{2005}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2006}\)

\(\Rightarrow3A-A=3^{2006}-3\)

\(\Rightarrow2A=3^{2006}-3\)

\(\Rightarrow2A+3=3^{2006}\) là 1 lũy thừa của 3 (đpcm)

4.

\(B=1+1+2+2^2+2^3+...+2^{100}\)

\(2B=2+2+2^2+...+2^{101}\)

\(\Rightarrow2B-B=2+2^{101}-\left(1+1\right)=2^{101}\)

\(\Rightarrow B=2^{101}\) là 1 lũy thừa của 2 (đpcm)

Bài 1:

\(A=2+2^2+2^3+...+2^{2003}+2^{2004}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2002}+2^{2003}+2^{2004}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2002}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+2^4+...+2^{2002}\right)⋮7\)

Bài 2:

\(A=2+2^2+2^3+2^4+...+2^{59}+2^{60}\)

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

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

\(=15\cdot\left(2+2^5+...+2^{57}\right)⋮15\)

Bài 3:

\(A=1+3+3^2+3^3+...+3^{1990}+3^{1991}\)

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

\(=13+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)

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

Bài 4:

\(A=4+4^2+4^3+4^4+...+4^{23}+4^{24}\)

\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{23}+4^{24}\right)\)

\(=\left(4+4^2\right)+4^2\left(4+4^2\right)+...+4^{22}\left(4+4^2\right)\)

\(=20\left(1+4^2+...+4^{22}\right)⋮20\)

bài 5,6,7,8,9 nữa ạaaa:33

\(a.\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\\ =\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{20}{21}\\ b.\dfrac{7}{12}-\dfrac{27}{7}\cdot\dfrac{1}{18}\\ =\dfrac{7}{12}-\dfrac{3}{14}=\dfrac{31}{84}\\ c.\left(\dfrac{23}{11}-\dfrac{15}{82}\right)\cdot\dfrac{41}{25}\\ =\dfrac{1721}{902}\cdot\dfrac{41}{25}=\dfrac{1721}{550}\\ d.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\\ =\dfrac{13}{10}\cdot\left(\dfrac{-5}{13}\right)=\dfrac{-1}{2}\)

Bài giải

Ta có: 6n + 4 \(⋮\)2n + 1   (n \(\inℤ\))

=> 6n + 4 - 3(2n + 1) \(⋮\)2n + 1

=> 1 \(⋮\)2n + 1

=> 2n + 1 \(\in\)Ư (1)

Ư (1) = {1; -1}

2n + 1 = 1 hay -1

2n       = 1 - 1 hay -1 - 1

2n       = 0 hay -2

  n       = 0 : 2 hay -2 : 2

  n       = 0 hay -1

Vậy n = 0 hay -1

Ư(240)={1;2;3;4;5;6;8;10;12;15;16;20;24;30;40;48;60;80;120;240}

Trong các số này thì các số là bội của 24 là:

24;48;120;240

Thank uu ><