Khi x > 9 thì 7 - x < 0 do đó |7-x| = x - 7 

\(\dfrac{3\left(x-1\right)}{4}+1< \dfrac{x+2}{3}\)

\(\Leftrightarrow\dfrac{9\left(x-1\right)}{12}+\dfrac{12}{12}< \dfrac{4\left(x+2\right)}{12}\)

\(\Rightarrow9\left(x-1\right)+12< 4\left(x+2\right)\)

\(\Leftrightarrow9x-9+12< 4x+8\)

\(\Leftrightarrow9x+3< 4x+8\)

\(\Leftrightarrow9x-4x< 8-3\)

\(\Leftrightarrow5x< 5\)

\(\Leftrightarrow x< 1\)

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 $4$ 

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

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

\(A\left(x\right)+B\left(x\right)=\left(3x^5-4x^3+2x^2-3\right)+\left(8x^4-x^3-9x+\dfrac{2}{5}\right)\)

\(=3x^5+8x^4+\left(-4x^3-x^3\right)+2x^2-9x+\left(-3+\dfrac{2}{5}\right)\)

\(=3x^5+8x^4-5x^3+2x^2-9x-\dfrac{13}{5}\)

\(A\left(x\right)-B\left(x\right)=\left(3x^5-4x^3+2x^2-3\right)-\left(8x^4-x^3-9x+\dfrac{2}{5}\right)\)

\(=3x^5-8x^4+\left(-4x^3+x^3\right)+2x^2+9x+\left(-3-\dfrac{2}{5}\right)\)

\(=3x^5-8x^4-3x^3+2x^2+9x-\dfrac{17}{5}\)

\(B\left(x\right)-A\left(x\right)=\left(8x^4-x^3-9x+\dfrac{2}{5}\right)-\left(3x^5-4x^3+2x^2-3\right)\)

\(=-3x^5+8x^4+\left(-x^3+4x^3\right)-2x^2-9x+\left(\dfrac{2}{5}+3\right)\)

\(=-3x^5+8x^4+3x^3-2x^2-9x+\dfrac{17}{5}\)