Đáp án:
\(
\frac{{10}}{{2x + 1}}
\)
Giải thích các bước giải:
\(
\begin{array}{l}
(\frac{{2x + 1}}{{2x - 1}} - \frac{{2x - 1}}{{2x + 1}}):\frac{{4x}}{{10x - 5}} \\
Đk:x \ne \pm \frac{1}{2} \\
= {\rm{[}}\frac{{(2x + 1)^2 - (2x - 1)^2 }}{{(2x - 1)(2x + 1)}}{\rm{]}}.\frac{{10x - 5}}{{4x}} \\
= \frac{{4x^2 + 4x + 1 - 4x^2 + 4x - 1}}{{(2x - 1)(2x + 1)}}.\frac{{5(2x - 1)}}{{4x}} \\
= \frac{{8x}}{{2x + 1}}.\frac{5}{{4x}} = \frac{{10}}{{2x + 1}} \\
\end{array}
\)
