Đáp án:
c) \(\left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\left| {1 - 2x} \right| = 3\\
\to \left[ \begin{array}{l}
1 - 2x = 3\\
1 - 2x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\left( l \right)\\
x = 2
\end{array} \right.\\
Thay:x = 2\\
\to A = \dfrac{3}{{4 + 2.2 + 1}} = \dfrac{3}{9} = \dfrac{1}{3}\\
2)B = \dfrac{{{x^2} - {{\left( {x + 1} \right)}^2}}}{{x{{\left( {x + 1} \right)}^2}}}\\
= \dfrac{{{x^2} - {x^2} - 2x - 1}}{{x{{\left( {x + 1} \right)}^2}}} = \dfrac{{ - 2x - 1}}{{x{{\left( {x + 1} \right)}^2}}}\\
P = A:B = \dfrac{3}{{{{\left( {x + 1} \right)}^2}}}:\dfrac{{ - 2x - 1}}{{x{{\left( {x + 1} \right)}^2}}}\\
= \dfrac{3}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{{x{{\left( {x + 1} \right)}^2}}}{{ - 2x - 1}}\\
= \dfrac{{3x}}{{ - 2x - 1}}\\
P = 2 \to \dfrac{{3x}}{{ - 2x - 1}} = 2\\
\to 3x = - 4x - 2\\
\to 7x = - 2\\
\to x = - \dfrac{2}{7}\\
3)P = \dfrac{{3x}}{{ - 2x - 1}} \to 2P = \dfrac{{6x}}{{ - 2x - 1}}\\
= \dfrac{{ - 3\left( { - 2x - 1} \right) - 3}}{{ - 2x - 1}}\\
= - 3 - \dfrac{3}{{ - 2x - 1}}\\
= - 3 + \dfrac{3}{{2x + 1}}\\
P \in Z \to \dfrac{3}{{2x + 1}} \in Z\\
\to 2x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
2x + 1 = 3\\
2x + 1 = - 3\\
2x + 1 = 1\\
2x + 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 2\\
x = 0\left( l \right)\\
x = - 1\left( l \right)
\end{array} \right.
\end{array}\)
