Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.DK:x \ne \left\{ { - \frac{3}{2};\frac{1}{2};\frac{5}{2};4} \right\}\\
P = \left[ {\frac{{2x - 3}}{{\left( {2x - 5} \right)\left( {2x - 1} \right)}} + \frac{{2x - 8}}{{\left( {4 - x} \right)\left( {2x - 5} \right)}} - \frac{3}{{2x - 1}}} \right]:\frac{{\left( {7 - 4x} \right)\left( {2x + 3} \right)}}{{\left( {2x - 1} \right)\left( {2x + 3} \right)}} + 1\\
= \frac{{ - 2{x^2} + 11x - 12 + 4{x^2} - 18x + 8 - 3\left( { - 2{x^2} + 13x - 20} \right)}}{{\left( {2x - 1} \right)\left( {4 - x} \right)\left( {2x - 5} \right)}}.\frac{{2x - 1}}{{7 - 4x}} + 1\\
= \frac{{2{x^2} - 7x - 4 + 6{x^2} - 39x + 60}}{{\left( {4 - x} \right)\left( {2x - 5} \right)\left( {7 - 4x} \right)}} + 1\\
= \frac{{2\left( {x - 4} \right)\left( {4x - 7} \right)}}{{\left( {4 - x} \right)\left( {2x - 5} \right)\left( {7 - 4x} \right)}} + 1\\
= \frac{{2 + 2x - 5}}{{2x - 5}} = \frac{{2x - 3}}{{2x - 5}}\\
b.\left| x \right| = \frac{1}{2} \to \left[ \begin{array}{l}
x = \frac{1}{2}\left( l \right)\\
x = - \frac{1}{2}
\end{array} \right.\\
\to A = \frac{2}{3}\\
c.A \in Z\\
\to \frac{{2x - 3}}{{2x - 5}} = \frac{{2x - 5 + 2}}{{2x - 5}} = 1 + \frac{2}{{2x - 5}} \in Z\\
\to \left( {2x - 5} \right) \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
2x - 5 = 2\\
2x - 5 = - 2\\
2x - 5 = 1\\
2x - 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \frac{7}{2}\left( l \right)\\
x = \frac{3}{2}\left( l \right)\\
x = 3\left( {TM} \right)\\
x = 2\left( {TM} \right)
\end{array} \right.\\
d.P > 0\\
\to \frac{{2x - 3}}{{2x - 5}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 3 > 0\\
2x - 5 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 3 < 0\\
2x - 5 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x > \frac{5}{2}\\
x < \frac{3}{2}
\end{array} \right.
\end{array}\)
