\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+\dfrac{14-11}{11.14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\Leftrightarrow x=\dfrac{1}{9}\)

x=0.1111111111

5/14a=3

=.>a=3:5/14

=>a=8.4

\(\dfrac{x-3}{10}-\dfrac{x-3}{15}=\dfrac{x-3}{12}-\dfrac{x-3}{18}\)

\(\Rightarrow\dfrac{x-3}{1}-\dfrac{x-3}{15}-\dfrac{x-3}{12}+\dfrac{x-3}{18}=0\)

\(\Rightarrow\left(x-3\right)\left(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\right)=0\)

Mà \(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\ne0\)

\(\Rightarrow x-3=0\Leftrightarrow x=3\)

\(Q=\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+\dfrac{70}{31.101}+\dfrac{90}{101.200}\)

\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\)

\(=\dfrac{1}{5}-\dfrac{1}{200}\)

\(=\dfrac{39}{200}\)

a. 

$10A=\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}<1+\frac{9}{10^{2020}+1}=\frac{10+10^{2020}}{10^{2020}+1}=10B$

$\Rightarrow A< B$

b.

\(C=2(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99})=2(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99})\)

\(=2(\frac{1}{3}-\frac{1}{99})=\frac{64}{99}\)

\(D=6(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{203.206})=6(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206})\)

$=6(\frac{1}{2}-\frac{1}{206})=\frac{306}{103}$

$C:D=\frac{64}{99}: \frac{306}{103}=\frac{3296}{15147}$

c.

\(A=\frac{12n}{3n+3}=\frac{12n}{3(n+1)}=\frac{4n}{n+1}=\frac{4(n+1)-4}{n+1}=4-\frac{4}{n+1}\)

Để $A$ nguyên thì $\frac{4}{n+1}$ nguyên 

$\Rightarrow n+1$ là ước của $4$ 

$\Rightarrow n+1\in\left\{\pm 1; \pm 4\right\}$

$\Rightarrow n\in\left\{0; -2; 3; -5\right\}$

Gọi \(d=ƯC\left(3n+15,n+4\right)\)

\(\Rightarrow\left\{{}\begin{matrix}3n+15⋮d\\n+4⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3n+15⋮d\\3n+12⋮d\end{matrix}\right.\)

\(\Rightarrow3⋮d\)

\(\Rightarrow d\inƯ\left(3\right)=\left\{1,3\right\}\)

Thay vào 
\(\Rightarrow d=1\)

Để 3n+15/n+4 là PSTG thì 3n+15 chia hết cho n+4

Mà n+4 cũng chia hết cho n+4 => 3(n+4) chia hết cho n+4 =>3n+12 chia hết cho n+4

=>(3n+12)-(3n+4) chia hết cho n+4

=>3n+12-3n-4 chia hết cho n+4

=>8 chia hết cho n+4

=>n+4 thuộc Ư(8)={1; -1 ;2; -2 ;4; -4 ;8; -8 } , ta có bản

Vậy n = {-3; -5; -2; -6; 0; -8; 4; -12}