Đáp án:
a. \(\dfrac{4}{{\left( {x - 2} \right)\left( {x - 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { - 2;1;2} \right\}\\
a.A = \left[ {\dfrac{1}{{x - 2}} - \dfrac{{2x}}{{4 - {x^2}}} + \dfrac{1}{{2 + x}}} \right].\dfrac{2}{{x - 1}}\\
= \left[ {\dfrac{{x + 2 + 2x - x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right].\dfrac{2}{{x - 1}}\\
= \dfrac{{2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{2}{{x - 1}}\\
= \dfrac{4}{{\left( {x - 2} \right)\left( {x - 1} \right)}}\\
b.2{x^2} + x = 0\\
\to x\left( {2x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{1}{2}
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = 0\\
x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \dfrac{4}{{\left( {0 - 2} \right)\left( {0 - 1} \right)}} = 2\\
A = \dfrac{4}{{\left( { - \dfrac{1}{2} - 2} \right)\left( { - \dfrac{1}{2} - 1} \right)}} = \dfrac{{16}}{{15}}
\end{array} \right.\\
c.A = \dfrac{1}{2}\\
\to \dfrac{4}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \dfrac{1}{2}\\
\to {x^2} - 3x + 2 = 8\\
\to {x^2} - 3x - 6 = 0\\
\to {x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{{33}}{4} = 0\\
\to {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{33}}{4}\\
\to \left[ \begin{array}{l}
x - \dfrac{3}{2} = \sqrt {\dfrac{{33}}{4}} \\
x - \dfrac{3}{2} = - \sqrt {\dfrac{{33}}{4}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{3 + \sqrt {33} }}{2}\\
x = \dfrac{{3 - \sqrt {33} }}{2}
\end{array} \right.\\
d.A = \dfrac{4}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \dfrac{4}{{{x^2} - 3x + 2}}\\
A \in {Z^ + }\\
\to \dfrac{4}{{{x^2} - 3x + 2}} \in {Z^ + }\\
\to {x^2} - 3x + 2 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
{x^2} - 3x + 2 = 4\\
{x^2} - 3x + 2 = 2\\
{x^2} - 3x + 2 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{3 \pm \sqrt {17} }}{2}\left( l \right)\\
x = 3\left( {TM} \right)\\
x = 0\left( {TM} \right)\\
x = \dfrac{{3 \pm \sqrt 5 }}{2}\left( l \right)
\end{array} \right.
\end{array}\)
