Đáp án:
d) \(\left[ \begin{array}{l}
x < 2;x \ne \left\{ { - 2;0} \right\}\\
x > \dfrac{9}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 2;0;2} \right\}\\
b)A = \left( {\dfrac{{{x^2}}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{6}{{3\left( {2 - x} \right)}} + \dfrac{x}{{x\left( {x + 2} \right)}}} \right):\left( {x - 2 + \dfrac{{10 - {x^2}}}{{x + 2}}} \right)\\
= \left[ {\dfrac{x}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{2}{{x - 2}} + \dfrac{1}{{x + 2}}} \right]:\left( {\dfrac{{{x^2} - 4 + 10 - {x^2}}}{{x + 2}}} \right)\\
= \dfrac{{x - 2x - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{6}\\
= \dfrac{{ - 6}}{{6\left( {x - 2} \right)}} = - \dfrac{1}{{x - 2}}\\
c)A = \dfrac{5}{6}\\
\to - \dfrac{1}{{x - 2}} = \dfrac{5}{6}\\
\to - 6 = 5x - 10\\
\to 5x = 4\\
\to x = \dfrac{4}{5}\\
d)A > - \dfrac{2}{5}\\
\to - \dfrac{1}{{x - 2}} > - \dfrac{2}{5}\\
\to \dfrac{1}{{x - 2}} < \dfrac{2}{5}\\
\to \dfrac{{5 - 2x + 4}}{{5\left( {x - 2} \right)}} < 0\\
\to \dfrac{{9 - 2x}}{{5\left( {x - 2} \right)}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
9 - 2x > 0\\
x - 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
9 - 2x < 0\\
x - 2 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{9}{2} > x\\
x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x > \dfrac{9}{2}\\
x > 2
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x < 2;x \ne \left\{ { - 2;0} \right\}\\
x > \dfrac{9}{2}
\end{array} \right.
\end{array}\)
