Đáp án:
b. \(\left[ \begin{array}{l}
x = \dfrac{{3 + \sqrt {17} }}{4}\\
x = \dfrac{{3 - \sqrt {17} }}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \left( {\dfrac{2}{{1 + 2x}} + \dfrac{{4{x^2} + 1}}{{4{x^2} - 1}} - \dfrac{1}{{1 - 2x}}} \right):\dfrac{2}{{4{x^2} - 1}}\\
= \left[ {\dfrac{{2\left( {1 - 2x} \right) + 4{x^2} + 1 - 1 - 2x}}{{\left( {1 + 2x} \right)\left( {1 - 2x} \right)}}} \right].\dfrac{{\left( {1 + 2x} \right)\left( {1 - 2x} \right)}}{2}\\
= \dfrac{{2 - 4x + 4{x^2} + 1 - 1 - 2x}}{{\left( {1 + 2x} \right)\left( {1 - 2x} \right)}}.\dfrac{{\left( {1 + 2x} \right)\left( {1 - 2x} \right)}}{2}\\
= \dfrac{{4{x^2} - 6x + 2}}{{\left( {1 + 2x} \right)\left( {1 - 2x} \right)}}.\dfrac{{\left( {1 + 2x} \right)\left( {1 - 2x} \right)}}{2}\\
= 2{x^2} - 3x + 1\\
b.A = 2\\
\to 2{x^2} - 3x + 1 = 2\\
\to 2{x^2} - 3x - 1 = 0\\
\to {\left( {x\sqrt 2 } \right)^2} - 2.x\sqrt 2 .\dfrac{3}{{2\sqrt 2 }} + {\left( {\dfrac{3}{{2\sqrt 2 }}} \right)^2} - \dfrac{{17}}{8} = 0\\
\to \left( {x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }}} \right) = \dfrac{{17}}{8}\\
\to \left[ \begin{array}{l}
x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }} = \dfrac{{17}}{8}\\
x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }} = - \dfrac{{17}}{8}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{3 + \sqrt {17} }}{4}\\
x = \dfrac{{3 - \sqrt {17} }}{4}
\end{array} \right.
\end{array}\)
