Đáp án:
\(-8.\)
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{2x + 1}}{{2{x^2} - x}} + \frac{{32{x^2}}}{{1 - 4{x^2}}} + \frac{{1 - 2x}}{{2{x^2} + x}}\\
= \frac{{2x + 1}}{{x\left( {2x - 1} \right)}} - \frac{{32{x^2}}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} - \frac{{2x - 1}}{{x\left( {2x + 1} \right)}}\\
= \frac{{{{\left( {2x + 1} \right)}^2} - 32{x^3} - {{\left( {2x - 1} \right)}^2}}}{{x\left( {2x - 1} \right)\left( {2x + 1} \right)}}\\
= \frac{{4{x^2} + 4x + 1 - 32{x^3} - 4{x^2} + 4x - 1}}{{x\left( {2x - 1} \right)\left( {2x + 1} \right)}}\\
= \frac{{ - 32{x^3} + 8x}}{{x\left( {4{x^2} - 1} \right)}} = \frac{{ - 8x\left( {4{x^2} - 1} \right)}}{{x\left( {4{x^2} - 1} \right)}} = - 8.
\end{array}\)
