Đáp án:
b. \(x = \frac{5}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left\{ \begin{array}{l}
4{x^2} - 1 \ne 0\\
4{x^2} + 4x + 1 \ne 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ne \frac{1}{2}\\
x \ne - \frac{1}{2}
\end{array} \right.\\
A = 1 - \left[ {\frac{{2\left( {1 - 2x} \right) + 5x - \left( {1 + 2x} \right)}}{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}} \right].\frac{{{{\left( {2x + 1} \right)}^2}}}{{x - 1}}\\
= 1 - \left[ {\frac{{2 - 4x + 5x - 1 - 2x}}{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}} \right].\frac{{{{\left( {2x + 1} \right)}^2}}}{{x - 1}}\\
= 1 - \left[ {\frac{{1 - x}}{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}} \right].\frac{{{{\left( {1 + 2x} \right)}^2}}}{{x - 1}}\\
= 1 + \left[ {\frac{{x - 1}}{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}} \right].\frac{{{{\left( {1 + 2x} \right)}^2}}}{{x - 1}}\\
= 1 + \frac{{1 + 2x}}{{1 - 2x}}\\
= \frac{{1 - 2x + 1 + 2x}}{{1 - 2x}}\\
= \frac{2}{{1 - 2x}}\\
b.A = - \frac{1}{2}\\
\to \frac{2}{{1 - 2x}} = - \frac{1}{2}\\
\to 4 = - 1 + 2x\\
\to 2x = 5\\
\to x = \frac{5}{2}\left( {TM} \right)
\end{array}\)
