Đáp án:

 B2:

e) \(A\max  = 4 \Leftrightarrow x =  - 4\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
a)DK:x \ne \left\{ { - 2;4} \right\}\\
b)B = \dfrac{{{x^2} - 3x + 14 + x\left( {x - 4} \right) + \left( {1 - x} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 4} \right)}}\\
 = \dfrac{{{x^2} - 3x + 14 + {x^2} - 4x - {x^2} - x + 2}}{{\left( {x + 2} \right)\left( {x - 4} \right)}}\\
 = \dfrac{{{x^2} - 8x + 16}}{{\left( {x + 2} \right)\left( {x - 4} \right)}} = \dfrac{{{{\left( {x - 4} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 4} \right)}} = \dfrac{{x - 4}}{{x + 2}}\\
c)Thay:x = 8\\
 \to B = \dfrac{{8 - 4}}{{8 + 2}} = \dfrac{4}{{10}} = \dfrac{2}{5}\\
d)M = A.B = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{2\left( {x - 4} \right)}}.\dfrac{{x - 4}}{{x + 2}} = \dfrac{{{x^2} + 2x + 1}}{{2x + 4}}\\
M = \dfrac{3}{4} \to \dfrac{{{x^2} + 2x + 1}}{{2x + 4}} = \dfrac{3}{4}\\
 \to 4{x^2} + 8x + 4 = 6x + 12\\
 \to 4{x^2} + 2x - 8 = 0\\
 \to \left( {4{x^2} + 2.2x.\dfrac{1}{2} + \dfrac{1}{4}} \right) - \dfrac{{33}}{4} = 0\\
 \to {\left( {2x + \dfrac{1}{2}} \right)^2} = \dfrac{{33}}{4}\\
 \to \left[ \begin{array}{l}
2x + \dfrac{1}{2} = \dfrac{{\sqrt {33} }}{2}\\
2x + \dfrac{1}{2} =  - \dfrac{{\sqrt {33} }}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt {33} }}{4}\\
x = \dfrac{{ - 1 - \sqrt {33} }}{4}
\end{array} \right.\\
e)M \le 0 \to \dfrac{{{x^2} + 2x + 1}}{{2x + 4}} \le 0\\
 \to 2x + 4 < 0\left( {do:{x^2} + 2x + 1 = {{\left( {x + 1} \right)}^2} \ge 0} \right)\\
 \to x <  - 2\\
B2:\\
a)DK:x \ne \left\{ { - 3; - 2;2} \right\}\\
b)B = \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right) + 2 - 3x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{{x^2} + x - 2 + 2 - 3x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{{x^2} - 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{x}{{x + 2}}\\
c)Thay:x = 3\\
 \to B = \dfrac{3}{{3 + 2}} = \dfrac{3}{5}\\
d)\dfrac{B}{A} = \dfrac{x}{{x + 2}}:\dfrac{x}{{x + 3}}\\
 = \dfrac{{x + 3}}{{x + 2}}\\
\dfrac{B}{A} \le 0 \to \dfrac{{x + 3}}{{x + 2}} \le 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 3 \ge 0\\
x + 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 3 \le 0\\
x + 2 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge  - 3\\
x <  - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le  - 3\\
x >  - 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
Do:x \ne 3\\
 \to  - 3 < x <  - 2\\
e)A = \dfrac{x}{{x + 3}} = \dfrac{{x + 3 - 3}}{{x + 3}}\\
 = 1 - \dfrac{3}{{x + 3}}\\
A \in Z \Leftrightarrow \dfrac{3}{{x + 3}} \in Z\\
 \Leftrightarrow x + 3 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
x + 3 = 3\\
x + 3 =  - 3\\
x + 3 = 1\\
x + 3 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 0\\
x =  - 6\\
x =  - 2\left( l \right)\\
x =  - 4
\end{array} \right. \to \left[ \begin{array}{l}
A = 0\\
A = 2\\
A = 4
\end{array} \right.\\
 \to A\max  = 4 \Leftrightarrow x =  - 4
\end{array}\)