Đáp án:
a) \(\dfrac{{x - 2}}{{3x}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
1)DK:x \ne \left\{ {0;1;2} \right\}\\
P = \left[ {\dfrac{{x - x + 1}}{{x\left( {x - 1} \right)}}} \right]:\left[ {\dfrac{{{x^2} - 1 - {x^2} + 4}}{{\left( {x - 2} \right)\left( {x - 1} \right)}}} \right]\\
= \dfrac{1}{{x\left( {x - 1} \right)}}.\dfrac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{3}\\
= \dfrac{{x - 2}}{{3x}}\\
2)\left| {x + 2} \right| = 3\\
\to \left[ \begin{array}{l}
x + 2 = 3\\
x + 2 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 5
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = 1\\
x = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
P = \dfrac{{1 - 2}}{{3.1}} = - \dfrac{1}{3}\\
P = \dfrac{{ - 5 - 2}}{{3\left( { - 5} \right)}} = \dfrac{7}{{15}}
\end{array} \right.\\
3)P > \dfrac{1}{3}\\
\to \dfrac{{x - 2}}{{3x}} > \dfrac{1}{3}\\
\to \dfrac{{3x - 6 - 3x}}{{9x}} > 0\\
\to - \dfrac{6}{{9x}} > 0\\
\to x < 0\\
4)P + \left| P \right| = 0\\
\to \left| P \right| = - P\\
\to \left[ \begin{array}{l}
P = - P\\
P = P\left( {ld} \right)
\end{array} \right.\\
\to 2P = 0\\
\to P = 0\\
\to \dfrac{{x - 2}}{{3x}} = 0\\
\to x = 2\left( l \right)\\
\to x \in \emptyset \\
5)3P = \dfrac{{3x - 6}}{{3x}} = 1 - \dfrac{2}{x}\\
P \in Z \to \dfrac{2}{x} \in Z\\
\to x \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = - 2\\
x = 1\left( l \right)\\
x = - 1
\end{array} \right.
\end{array}\)
