Đáp án:

$\begin{array}{l}
a)B = \left( {\dfrac{{9 - 3x}}{{{x^2} + 4x - 5}} - \dfrac{{x + 5}}{{1 - x}} - \dfrac{{x + 1}}{{x + 5}}} \right)\\
:\dfrac{{7x - 14}}{{{x^3} - 1}}\\
 = \dfrac{{9 - 3x + \left( {x + 5} \right)\left( {x + 5} \right) - \left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 5} \right)}}\\
.\dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{7\left( {x - 2} \right)}}\\
 = \dfrac{{9 - 3x + {x^2} + 10x + 25 - {x^2} + 1}}{{x + 5}}.\dfrac{{{x^2} + x + 1}}{{7\left( {x - 2} \right)}}\\
 = \dfrac{{7x + 35}}{{x + 5}}.\dfrac{{{x^2} + x + 1}}{{7\left( {x - 2} \right)}}\\
 = \dfrac{{{x^2} + x + 1}}{{x - 2}}\\
b){\left( {x + 5} \right)^2} - 9x - 45 = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {x + 5 - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {x - 4} \right) = 0\\
 \Leftrightarrow x =  - 5\left( {ktm} \right);x = 4\left( {tm} \right)\\
 + Khi:x = 4\\
 \Leftrightarrow B = \dfrac{{{4^2} + 4 + 1}}{{4 - 2}} = \dfrac{{21}}{2}\\
c)B = \dfrac{{{x^2} + x + 1}}{{x - 2}}\\
 = \dfrac{{{x^2} - 4 + x - 2 + 7}}{{x - 2}}\\
 = x + 2 + 1 + \dfrac{7}{{x - 2}}\\
 = x + 3 + \dfrac{7}{{x - 2}}\\
B \in Z\\
 \Leftrightarrow \dfrac{7}{{x - 2}} \in Z\\
 \Leftrightarrow \left( {x - 2} \right) \in \left\{ { - 7; - 1;1;7} \right\}\\
 \Leftrightarrow x \in \left\{ { - 5;1;3;9} \right\}\\
Do:x\#  - 5;x\# 1\\
 \Leftrightarrow x \in \left\{ {3;9} \right\}\\
d)B = \dfrac{{ - 3}}{4}\\
 \Leftrightarrow \dfrac{{{x^2} + x + 1}}{{x - 2}} =  - \dfrac{3}{4}\\
 \Leftrightarrow 4{x^2} + 4x + 4 =  - 3x + 6\\
 \Leftrightarrow 4{x^2} + 7x - 2 = 0\\
 \Leftrightarrow \left( {4x - 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow x = \dfrac{1}{4};x =  - 2\left( {tm} \right)\\
Vậy\,x = \dfrac{1}{4};x =  - 2
\end{array}$

$\begin{array}{l}
e)B < 0\\
 \Leftrightarrow \dfrac{{{x^2} + x + 1}}{{x - 2}} < 0\\
 \Leftrightarrow x - 2 < 0\\
 \Leftrightarrow x < 2\\
Vậy\,x < 2;x\# 1;x\#  - 5\\
f)M = \dfrac{2}{{x - 2}}:B\\
 = \dfrac{2}{{x - 2}}.\dfrac{{x - 2}}{{{x^2} + x + 1}}\\
 = \dfrac{2}{{{x^2} + x + 1}}\\
Do:{x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
 \Leftrightarrow \dfrac{1}{{{x^2} + x + 1}} \le \dfrac{4}{3}\\
 \Leftrightarrow \dfrac{2}{{{x^2} + x + 1}} \le \dfrac{8}{3}\\
 \Leftrightarrow GTLN:M = \dfrac{8}{3}\,khi:x =  - \dfrac{1}{2}\\
g)B = \dfrac{{{x^2} + x + 1}}{{x - 2}}\\
 = x + 3 + \dfrac{7}{{x - 2}}\\
 = x - 2 + \dfrac{7}{{x - 2}} + 5\\
Theo\,Co - si:\\
\left( {x - 2} \right) + \dfrac{7}{{x - 2}} \ge 2\sqrt {\left( {x - 2} \right).\dfrac{7}{{x - 2}}}  = 2\sqrt 7 \\
 \Leftrightarrow \left( {x - 2} \right) + \dfrac{7}{{x - 2}} + 5 \ge 2\sqrt 7  + 5\\
 \Leftrightarrow B \ge 2\sqrt 7  + 5\\
 \Leftrightarrow GTNN:B = 2\sqrt 7  + 5
\end{array}$