ĐK: \(x\ne\pm2\)
\( a)A = \left( {\dfrac{3}{{2x + 4}} + \dfrac{x}{{2 - x}} + \dfrac{{2{x^2} + 3}}{{{x^2} - 4}}} \right):\left( {\dfrac{{2x - 1}}{{4x - 8}}} \right)\\ = \dfrac{{3\left( {2 - x} \right) + 2x\left( {2 + x} \right) - 4{x^2} - 6}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}:\dfrac{{2x - 1}}{{4\left( {x - 2} \right)}}\\ = \dfrac{{ - 2{x^2} + x}}{{2\left( {x + 2} \right)}}.\dfrac{{4\left( {x - 2} \right)}}{{2x - 1}} = \dfrac{{2x}}{{x + 2}} \)
\(b)\dfrac{{2x}}{{x + 2}} - 2 = \dfrac{{2x - 2x + 4}}{{x + 2}} = \dfrac{4}{{x + 2}}\)
Ta có: \(4>0\) để \(\dfrac{4}{{x + 2}} < 0 \Rightarrow x + 2 < 0 \Rightarrow x < - 2\)
\(c)\left| {x - 1} \right| = 3 \Leftrightarrow \left[ \begin{array}{l} x - 1 = 3\\ x - 1 = - 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 4\left( {tm} \right)\\ x = - 2\left( {ktm} \right) \end{array} \right.\)
Với \(x = 4 \Rightarrow A = \dfrac{{2.4}}{{4 + 2}} = \dfrac{4}{3}\)
\( d)\left| A \right| = \left| {\dfrac{{2x}}{{x + 2}}} \right| = 1\\ T{H_1}:\dfrac{{2x}}{{x + 2}} = 1 \Leftrightarrow 2x = x + 2 \Leftrightarrow x = 2\\ T{H_2}:\dfrac{{2x}}{{x + 2}} = - 1 \Leftrightarrow 2x = - x - 2 \Leftrightarrow x = - \dfrac{2}{3} \)
