Đáp án:
$\frac{{3\left( {2x + 1} \right)}}{{2x}}$
Giải thích các bước giải:
\(A = \frac{3}{{2x}} + \frac{{3x - 3}}{{2x - 1}} + \frac{{2{x^2} + 1}}{{4{x^2} - 2x}}\)
Điều kiện: \(\left\{ \begin{array}{l}2x \ne 0\\2x - 1 \ne 0\\4{x^2} - 2x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 0\\2x \ne 1\\2x\left( {2x - 1} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 0\\x \ne \frac{1}{2}\end{array} \right..\)
\(\begin{array}{l}A = \frac{3}{{2x}} + \frac{{3x - 3}}{{2x - 1}} + \frac{{2{x^2} + 1}}{{4{x^2} - 2x}}\\ = \frac{3}{{2x}} + \frac{{3x - 3}}{{2x - 1}} + \frac{{2{x^2} + 1}}{{2x\left( {2x - 1} \right)}}\\ = \frac{{3\left( {2x - 1} \right) + \left( {3x - 3} \right).2x + 2{x^2} + 1}}{{2x\left( {2x - 1} \right)}}\\ = \frac{{6x - 3 + 6{x^2} - 6x + 2{x^2} + 1}}{{2x\left( {2x - 1} \right)}}\\ = \frac{{8{x^2} - 2}}{{2x\left( {2x - 1} \right)}} = \frac{{3\left( {4{x^2} - 1} \right)}}{{2x\left( {2x - 1} \right)}}\\ = \frac{{3\left( {2x - 1} \right)\left( {2x + 1} \right)}}{{2x\left( {2x - 1} \right)}} = \frac{{3\left( {2x + 1} \right)}}{{2x}}.\end{array}\)
