Đáp án:
c) \(x = \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 3; - 1;1} \right\}\\
b)D = \left[ {\dfrac{{2x + 2 - x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right].\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{{{\left( {x + 3} \right)}^2}}} + \dfrac{{x + 1}}{{2\left( {x + 3} \right)}}\\
= \dfrac{{x + 3}}{{{{\left( {x + 3} \right)}^2}}} + \dfrac{{x + 1}}{{2\left( {x + 3} \right)}}\\
= \dfrac{1}{{x + 3}} + \dfrac{{x + 1}}{{2\left( {x + 3} \right)}} = \dfrac{{2 + x + 1}}{{2\left( {x + 3} \right)}}\\
= \dfrac{{x + 3}}{{2\left( {x + 3} \right)}} = \dfrac{1}{2}\\
c)D\left( x \right) = - 2{x^2} + 2x\\
\to - 2{x^2} + 2x = \dfrac{1}{2}\\
\to - 4{x^2} + 4x - 1 = 0\\
\to - {\left( {2x - 1} \right)^2} = 0\\
\to 2x - 1 = 0\\
\to x = \dfrac{1}{2}
\end{array}\)
