Đáp án:
a) \(\dfrac{{2x - 3}}{{2x - 5}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - \dfrac{3}{2};\dfrac{1}{2};\dfrac{5}{2};4} \right\}\\
P = \left[ {\dfrac{{2x - 3}}{{\left( {2x - 1} \right)\left( {2x - 5} \right)}} + \dfrac{{2x - 8}}{{\left( {4 - x} \right)\left( {2x - 5} \right)}} - \dfrac{3}{{2x - 1}}} \right]:\dfrac{{\left( {7 - 4x} \right)\left( {2x + 3} \right)}}{{\left( {2x - 1} \right)\left( {2x + 3} \right)}} + 1\\
= \left[ {\dfrac{{2x - 3}}{{\left( {2x - 1} \right)\left( {2x - 5} \right)}} - \dfrac{2}{{2x - 5}} - \dfrac{3}{{2x - 1}}} \right].\dfrac{{2x - 1}}{{7 - 4x}} + 1\\
= \dfrac{{2x - 3 - 2\left( {2x - 1} \right) - 3\left( {2x - 5} \right)}}{{\left( {2x - 1} \right)\left( {2x - 5} \right)}}.\dfrac{{2x - 1}}{{7 - 4x}} + 1\\
= \dfrac{{ - 8x + 14}}{{\left( {2x - 1} \right)\left( {2x - 5} \right)}}.\dfrac{{2x - 1}}{{7 - 4x}} + 1\\
= \dfrac{{2\left( { - 4x + 7} \right)}}{{\left( {2x - 1} \right)\left( {2x - 5} \right)}}.\dfrac{{2x - 1}}{{7 - 4x}} + 1\\
= \dfrac{2}{{2x - 5}} + 1 = \dfrac{{2 + 2x - 5}}{{2x - 5}} = \dfrac{{2x - 3}}{{2x - 5}}\\
b)\left| x \right| = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\left( l \right)\\
x = - \dfrac{1}{2}
\end{array} \right.\\
Thay:x = - \dfrac{1}{2}\\
\to P = \dfrac{{2.\left( { - \dfrac{1}{2}} \right) - 3}}{{2\left( { - \dfrac{1}{2}} \right) - 5}} = \dfrac{2}{3}\\
c)P = \dfrac{{2x - 3}}{{2x - 5}} = \dfrac{{2x - 5 + 2}}{{2x - 5}} = 1 + \dfrac{2}{{2x - 5}}\\
P \in Z \to \dfrac{2}{{2x - 5}} \in Z\\
\to 2x - 5 \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 = \dfrac{7}{2}\left( l \right)\\
x = \dfrac{3}{2}\left( l \right)\\
x = 3\left( {TM} \right)\\
x = 2\left( {TM} \right)
\end{array} \right.\\
d)P > 0\\
\to \dfrac{{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 > \dfrac{5}{2}\\
x < \dfrac{3}{2}
\end{array} \right.
\end{array}\)