Đáp án:

b) \(\dfrac{{6x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
a)DK:x \ne \left\{ { - 4;1;4} \right\}\\
\dfrac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 4} \right)}} - \dfrac{3}{{x - 1}} + \dfrac{4}{{x - 4}}\\
 = \dfrac{{\left( {2x - 3} \right)\left( {x - 4} \right) - 3\left( {x + 4} \right)\left( {x - 4} \right) + 4\left( {x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 1} \right)\left( {x + 4} \right)\left( {x - 4} \right)}}\\
 = \dfrac{{2{x^2} - 11x + 12 - 3\left( {{x^2} - 16} \right) + 4\left( {{x^2} + 3x - 4} \right)}}{{\left( {x - 1} \right)\left( {x + 4} \right)\left( {x - 4} \right)}}\\
 = \dfrac{{2{x^2} - 11x + 12 - 3{x^2} + 42 + 4{x^2} + 12x - 16}}{{\left( {x - 1} \right)\left( {x + 4} \right)\left( {x - 4} \right)}}\\
 = \dfrac{{3{x^2} + x + 38}}{{\left( {x - 1} \right)\left( {x + 4} \right)\left( {x - 4} \right)}}\\
b)DK:x \ne  \pm 1\\
\dfrac{{2{x^2} - 1 - 5x\left( {x - 1} \right) + \left( {3x - 2} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2{x^2} - 1 - 5{x^2} + 5x + 3{x^2} + x - 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{6x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}
\end{array}\)