Đáp án:
\[P = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left| {4x + 3} \right| - \left| {x - 1} \right| < x\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,x < - \frac{3}{4} \Rightarrow \left\{ \begin{array}{l}
4x + 3 < 0\\
x - 1 < 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow - \left( {4x + 3} \right) + \left( {x - 1} \right) < x\\
\Leftrightarrow - 4x - 4 < 0\\
\Leftrightarrow x > - 1\\
\Rightarrow {S_1} = \left( { - 1; - \frac{3}{4}} \right)\\
TH2:\,\,\, - \frac{3}{4} \le x \le 1 \Rightarrow \left\{ \begin{array}{l}
4x + 3 \ge 0\\
x - 1 \le 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {4x + 3} \right) + \left( {x - 1} \right) < x\\
\Leftrightarrow 4x < - 2\\
\Leftrightarrow x < - \frac{1}{2}\\
\Rightarrow {S_2} = \left[ { - \frac{3}{4}; - \frac{1}{2}} \right)\\
TH3:\,\,\,x > 1 \Rightarrow \left\{ \begin{array}{l}
4x + 3 > 0\\
x - 1 > 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {4x + 3} \right) - \left( {x - 1} \right) < x\\
\Leftrightarrow 2x < - 4\\
\Leftrightarrow x < - 2\\
\Rightarrow {S_3} = \emptyset \\
S = {S_1} \cup {S_2} \cup {S_3} = \left( { - 1; - \frac{1}{2}} \right)\\
P = 2a - 4b = - 2 + 2 = 0
\end{array}\)
