Đáp án:

\(\begin{array}{l}
B5:\\
a)\dfrac{x}{{x + 3}}\\
b)\left[ \begin{array}{l}
A = \dfrac{5}{8}\\
A =  - \dfrac{1}{2}
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 0\\
x =  - 6\\
x =  - 2\\
x =  - 4
\end{array} \right.\\
d)MinA =  - 2\\
e)x =  - 2\\
B6:\\
a)\dfrac{1}{{3x - 2}}\\
b)x \in \emptyset 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B5:\\
a)DK:x \ne 1\\
A = \dfrac{{2{x^2} + 1 - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}:\dfrac{{{x^2} + x + 1 - {x^2} + 2}}{{{x^2} + x + 1}}\\
 = \dfrac{{{x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 3}}\\
 = \dfrac{{x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 3}}\\
 = \dfrac{x}{{x + 3}}\\
b)\left| {x - 2} \right| = 3\\
 \to \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 =  - 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  - 1
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = 5\\
x =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
A = \dfrac{5}{{5 + 3}} = \dfrac{5}{8}\\
A = \dfrac{{ - 1}}{{ - 1 + 3}} =  - \dfrac{1}{2}
\end{array} \right.\\
c)A = \dfrac{x}{{x + 3}} = \dfrac{{x + 3 - 3}}{{x + 3}}\\
 = 1 - \dfrac{3}{{x + 3}}\\
A \in Z \to \dfrac{3}{{x + 3}} \in Z\\
 \to 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\\
x =  - 4
\end{array} \right.\\
d)A = 1 - \dfrac{3}{{x + 3}}\\
A\min  \Leftrightarrow \dfrac{3}{{x + 3}}\max \\
 \Leftrightarrow \left( {x + 3} \right)\min \\
 \to x + 3 = 1\\
 \to x =  - 2\\
 \to MinA = 1 - \dfrac{3}{{ - 2 + 3}} =  - 2\\
e)A =  - 2\\
 \to \dfrac{x}{{x + 3}} =  - 2\\
 \to x =  - 2x - 6\\
 \to 3x =  - 6\\
 \to x =  - 2\\
B6:\\
a)DK:x \ne \left\{ {1;\dfrac{3}{2}} \right\}\\
C = \dfrac{{2x - 5\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {2x - 3} \right)}}:\dfrac{{3 - 3x + 2}}{{1 - x}}\\
 = \dfrac{{2x - 5x + 5}}{{\left( {x - 1} \right)\left( {2x - 3} \right)}}.\dfrac{{1 - x}}{{5 - 3x}}\\
 = \dfrac{{5 - 3x}}{{\left( {x - 1} \right)\left( {2x - 3} \right)}}.\dfrac{{1 - x}}{{5 - 3x}}\\
 =  - \dfrac{1}{{2x - 3}} = \dfrac{1}{{3x - 2}}\\
b)C\min \\
 \Leftrightarrow \left( {\dfrac{1}{{3x - 2}}} \right)\min \\
 \Leftrightarrow \left( {3x - 2} \right)\max \\
 \to x \in \emptyset 
\end{array}\)