Lời giải:
a. 

$\frac{3n+2}{3}=n+\frac{2}{3}> n+\frac{1}{2}=\frac{2n+1}{2}$

$\Rightarrow \frac{3}{3n+2}< \frac{2}{2n+1}$
b.

$\frac{2n+1}{2n}=1+\frac{1}{2n}> 1+\frac{1}{3n}=\frac{1+3n}{3n}$

$\Rightarrow \frac{2n}{2n+1}< \frac{3n}{3n+1}$

\(\dfrac{3}{7}-\dfrac{1}{2}x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{7}-\dfrac{5}{3}\)

\(\Rightarrow x=2\cdot-\dfrac{26}{21}\)

\(\Rightarrow x=\dfrac{-52}{21}\) 

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\(2x-\dfrac{3}{4}=\dfrac{-5}{8}\)

\(\Rightarrow2x=\dfrac{-5}{8}+\dfrac{3}{4}\)

\(\Rightarrow2x=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1}{8}:2\)

\(\Rightarrow x=\dfrac{1}{16}\) 

________

\(\dfrac{1}{4}x-\left(-\dfrac{7}{5}\right)=\dfrac{-5}{3}\)

\(\Rightarrow\dfrac{1}{4}x+\dfrac{7}{5}=\dfrac{-5}{3}\)

\(\Rightarrow\dfrac{1}{4}x=\dfrac{-5}{3}-\dfrac{7}{5}\)

\(\Rightarrow x=4\cdot-\dfrac{46}{15}\)

\(\Rightarrow x=-\dfrac{184}{15}\)

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\(2\dfrac{1}{3}x-\dfrac{3}{4}=1\dfrac{1}{6}\)

\(\Rightarrow\dfrac{7}{3}x-\dfrac{3}{4}=\dfrac{7}{6}\)

\(\Rightarrow\dfrac{7}{3}x=\dfrac{7}{6}+\dfrac{3}{4}\)

\(\Rightarrow\dfrac{7}{3}x=\dfrac{23}{12}\)

\(\Rightarrow x=\dfrac{23}{12}:\dfrac{7}{3}\)

\(\Rightarrow x=\dfrac{23}{28}\) 

Bạn làm nốt cho mình bài trên dc ko

\(\dfrac{-5}{8}:\dfrac{15}{4}=\dfrac{-5}{8}\cdot\dfrac{4}{15}=\dfrac{-1\cdot1}{2\cdot3}=\dfrac{-1}{6}\)

\(\dfrac{-15}{17}:\dfrac{25}{-34}=\dfrac{-15}{17}\cdot\dfrac{-34}{25}=\dfrac{-3\cdot-2}{1\cdot5}=\dfrac{-6}{5}\)

\(-12:\dfrac{8}{3}=-12\cdot\dfrac{3}{8}=\dfrac{-12\cdot3}{8}=\dfrac{-3\cdot3}{2}=\dfrac{-9}{2}\)

\(\dfrac{-15}{14}:\dfrac{20}{-21}=\dfrac{-15}{14}\cdot\dfrac{-21}{20}=\dfrac{-3\cdot-3}{2\cdot4}=\dfrac{-9}{8}\)

\(-48:\dfrac{-24}{5}=-48\cdot\dfrac{5}{-24}=\dfrac{-48\cdot5}{-24}=2\cdot5=10\) 

\(\dfrac{-30}{7}:\dfrac{-5}{-14}=\dfrac{-30}{7}:\dfrac{5}{14}=\dfrac{-30}{7}\cdot\dfrac{14}{5}=\dfrac{-6\cdot2}{1\cdot1}=-18\) 

a) Gọi \(ƯCLN\left(a^2,a+b\right)=d\)  với \(d\inℕ^∗\)

\(\Rightarrow\left\{{}\begin{matrix}a^2⋮d\\a+b⋮d\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}a^2⋮d\\a^2+ab⋮d\end{matrix}\right.\) 

\(\Rightarrow ab⋮d\) 

Vì \(a,b\) nguyên tố cùng nhau \(\Rightarrow\left[{}\begin{matrix}a⋮d\\b⋮d\end{matrix}\right.\)

Hơn nữa, vì \(a+b⋮d\) nên nếu \(a⋮d\) thì \(b⋮d\). Nếu \(b⋮d\) thì \(a⋮d\). Như vậy \(a,b⋮d\).

 Nhưng do \(a,b\) nguyên tố cùng nhau nên \(d=1\) \(\RightarrowƯCLN\left(a^2,a+b\right)=1\) hay \(a^2,a+b\) nguyên tố cùng nhau.

b) Gọi \(ƯCLN\left(ab,a+b\right)=d\)

\(\Rightarrow\left\{{}\begin{matrix}ab⋮d\\a+b⋮d\end{matrix}\right.\)

 Vì a và b nguyên tố cùng nhau nên từ \(ab⋮d\Rightarrow\left[{}\begin{matrix}a⋮d\\b⋮d\end{matrix}\right.\). Đến đây kết hợp với \(a+b⋮d\)  và lập luận tương tự như câu a), sẽ chứng minh được \(d=1\)

Bài 12:

a) \(\dfrac{12}{3n-1}\) là một số nguyên khi:

\(12\) ⋮ \(3n-1\) 

\(\Rightarrow3n-1\inƯ\left(12\right)=\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

\(\Rightarrow3n\in\left\{2;0;3;-1;4;-2;5;-3;7;-5;13;-11\right\}\)

\(\Rightarrow n\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3};\dfrac{4}{3};-\dfrac{2}{3};\dfrac{5}{3};-1;\dfrac{7}{3};-\dfrac{5}{3};\dfrac{13}{3};-\dfrac{11}{3}\right\}\)

Mà: \(n\in Z\Rightarrow n\in\left\{0;1;-1\right\}\)

b) \(\dfrac{2n+3}{7}\) là một số nguyên khi:

\(2n+3\) ⋮ 7 

\(\Rightarrow2n+3\in B\left(7\right)\) 

Do \(n\in Z\) nên \(2n+3\) lẻ 

\(\Rightarrow2n+3\in B\left(7,\text{lẻ}\right)\)

\(\Rightarrow n\in\dfrac{B\left(7,\text{lẻ}\right)-3}{2}\) 

c) \(\dfrac{2n+5}{n-3}=\dfrac{2n-6+11}{n-3}=\dfrac{2\left(n-3\right)+11}{n-3}=2+\dfrac{11}{n-3}\)

Là một số nguyên khi:

11 ⋮ \(n-3\)

\(\Rightarrow n-3\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow n\in\left\{4;2;14;-8\right\}\)

Bài 8: 

S = \(\dfrac{3}{10}\) + \(\dfrac{3}{11}\) + \(\dfrac{3}{12}\) + \(\dfrac{3}{13}\) + \(\dfrac{3}{14}\)

\(\dfrac{3}{10}>\dfrac{3}{11}>\dfrac{3}{12}>\dfrac{3}{13}>\dfrac{3}{14}\)

S < \(\dfrac{3}{10}\).5 = \(\dfrac{3}{2}\) = 1 + \(\dfrac{1}{2}\) 

S > \(\dfrac{3}{14}.5\) = \(\dfrac{15}{14}\) =  1 + \(\dfrac{1}{14}\) 

Vì \(\dfrac{1}{14}\) < \(\dfrac{1}{2}\) nên 

1 < \(\dfrac{15}{14}\) < S < \(\dfrac{3}{2}\) < 2 

Vậy 1 < S < 2 (đpcm)

2008 nhé chúc năm mới vui vẻ

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2009}\)

\(x+1=2009\)

\(x=2008\)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dots+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dots+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2008}{2009}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Rightarrow x=2009-1\)

\(\Rightarrow x=2008\)

\(\dfrac{3}{7}\cdot\dfrac{5}{4}-\dfrac{5}{7}\cdot\dfrac{1}{3}-\dfrac{1}{7}\cdot\dfrac{5}{12}\)

\(=\dfrac{15}{28}-\dfrac{5}{21}-\dfrac{5}{84}\)

\(=\dfrac{25}{84}-\dfrac{5}{84}\)

\(=\dfrac{5}{21}\)

5/21  chúc năm mới vui vẻ

\(\dfrac{4}{7}x-x=-\dfrac{9}{14}\)

\(\left(\dfrac{4}{7}-1\right)x=-\dfrac{9}{14}\)

\(-\dfrac{3}{7}x=-\dfrac{9}{14}\)

\(x=-\dfrac{9}{14}:\dfrac{-3}{7}\)

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

\(\dfrac{4}{7}x-x=-\dfrac{9}{14}\)

\(\Rightarrow x\cdot\left(\dfrac{4}{7}-1\right)=-\dfrac{9}{14}\)

\(\Rightarrow x\cdot\dfrac{-3}{7}=\dfrac{-9}{14}\)

\(\Rightarrow x=\dfrac{-9}{14}:\dfrac{-3}{7}\)

\(\Rightarrow x=\dfrac{3}{2}\)

Bạn chỉnh lại môn học nhé.

Bài 2:

d; \(\dfrac{3}{5}\) : (-5) = \(\dfrac{3}{5}\) x \(\dfrac{1}{\left(-5\right)}\) = \(\dfrac{3}{-25}\) = \(\dfrac{-3}{25}\)

e; \(\dfrac{-4}{15}\): 2 = \(\dfrac{-4}{15}\) x \(\dfrac{1}{2}\) = \(\dfrac{-4}{30}\) 

f; 24 : \(\dfrac{-6}{7}\) = 24 x \(\dfrac{7}{-6}\) = \(\dfrac{28}{-1}\) = -28

Bài 3:

b; \(\dfrac{-4}{5}\) : \(\dfrac{2}{7}\) + \(\dfrac{4}{7}\)

= \(\dfrac{-4}{5}.\dfrac{7}{2}\) + \(\dfrac{4}{7}\)

= \(\dfrac{-14}{5}\) + \(\dfrac{4}{7}\)

= \(\dfrac{-98}{35}\) + \(\dfrac{20}{35}\)

= \(\dfrac{-78}{35}\)