a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3

\(\Rightarrow\) n - 2 \(\in\) Ư(3)

\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}

n \(\in\){5; -1; 3; 2}

c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{3}-\dfrac{1}{30}\)

\(=\dfrac{10}{30}-\dfrac{1}{30}\)

\(=\dfrac{9}{30}\)

=\(\dfrac{3}{10}\)

\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)

\(\left(x+1\right)^2=324=18^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=18\\x+1=-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-19\end{matrix}\right.\)

Ta có \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{x+1}{324}+\dfrac{1}{x+1}\)

\(\Rightarrow\)\(\dfrac{1}{x+1}-\dfrac{x+1}{324}=0\)

\(\Rightarrow\)\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)

\(\Rightarrow\)(x+1).(x+1)=324

\(\Rightarrow\)(x+1)²=324

\(\Rightarrow\)(x+1)² = 18² = (-18)²

TH1: (x+1)² = 18²

\(\Rightarrow\)x+1 = 18 

\(\Rightarrow\)x = 17

TH2: (x+1)² = (-18)²

\(\Rightarrow\)x+1 = -18 

\(\Rightarrow\)x = -19

Vậy x\(\in\)\(\left\{17;-19\right\}\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{600}\)

=> x + 1 = 600

x = 600 - 1

x = 599

Vậy x = 599

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{3}\) .-.

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

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

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{200.201}\)

=\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{200}-\dfrac{1}{201}\)

=\(\dfrac{1}{3}-\dfrac{1}{201}\)

=\(\dfrac{67}{201}-\dfrac{1}{201}\)

=\(\dfrac{66}{201}\)

                         ---Học Tốt Nha---

sai rùi

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n+1\right)}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)

= 1 - \(\dfrac{1}{n+1}\) = \(\dfrac{n}{n+1}\)

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

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

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

\(M=1-\dfrac{1}{7}\)

\(M=\dfrac{6}{7}\)

tham khảo

https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5

vào đi 

11: \(=\left(1+\dfrac{1}{98}-1-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\right)=0\)

12: \(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\left(\dfrac{-6+5}{10}\right)^2\)

\(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\dfrac{1}{100}=\dfrac{7}{17}+\dfrac{1}{170}=\dfrac{71}{170}\)