Giải thích các bước giải:
\(
\begin{array}{l}
a)(\frac{{2x + 1}}{{2x - 1}} - \frac{{2x - 1}}{{2x + 1}}):\frac{{4x}}{{10x - 5}} \\
dk:x \ne \pm \frac{1}{2} \\
= \frac{{(2x + 1)^2 - (2x - 1)^2 }}{{(2x - 1)(2x + 1)}}.\frac{{10x - 5}}{{4x}} \\
= \frac{{4x^2 + 4x + 1 - 4x^2 + 4x - 1}}{{(2x - 1)(2x + 1)}}.\frac{{5(2x - 1)}}{{4x}} \\
= \frac{{8x}}{{2x + 1}}.\frac{5}{{4x}} = \frac{{10}}{{2x + 1}} \\
b)(\frac{{x + 2}}{{x + 1}} - \frac{{2x}}{{x - 1}}).\frac{{3x + 3}}{x} + \frac{{4x^2 + x + 7}}{{x^2 - x}} \\
dk:\left\{ {\begin{array}{*{20}c}
\begin{array}{l}
x \ne 0 \\
x \ne 1 \\
\end{array} \\
{x \ne - 1} \\
\end{array}} \right. \\
= \frac{{(x + 2)(x - 1) - 2x(x + 1)}}{{(x - 1)(x + 1)}}.\frac{{3(x + 1)}}{x} + \frac{{4x^2 + x + 7}}{{x(x - 1)}} \\
= \frac{{x^2 + x - 2 - 2x^2 - 2x}}{{x - 1}}.\frac{3}{x} + \frac{{4x^2 + x + 7}}{{x(x - 1)}} \\
= \frac{{ - x^2 - x - 2}}{{x - 1}}.\frac{3}{x} + \frac{{4x^2 + x + 7}}{{x(x - 1)}} \\
= \frac{{ - 3x^2 - 3x - 6 + 4x^2 + x + 7}}{{x(x - 1)}} \\
= \frac{{x^2 - 2x + 1}}{{x(x - 1)}} \\
= \frac{{(x - 1)^2 }}{{x(x - 1)}} = \frac{{x - 1}}{x} \\
\end{array}
\)
