Đáp án:
\(\begin{array}{l}
a) - \dfrac{4}{{x + 2}}\\
b)A = - 2\\
c)x = - \dfrac{{10}}{3}\\
d)\left[ \begin{array}{l}
x > - 2\\
x < - 6
\end{array} \right.\\
e)\left[ \begin{array}{l}
x = 6\\
x = - 10\\
x = - 6\\
x = 0\\
x = - 4\\
x = - 1\\
x = - 3
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left[ {\dfrac{x}{{x + 2}} + \dfrac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}.\dfrac{{{x^2} - 2x + 4}}{{\left( {2 - x} \right)\left( {x + 2} \right)}}} \right].\dfrac{{x + 2}}{4}\\
= \left[ {\dfrac{x}{{x + 2}} - \dfrac{{{x^2} + 2x + 4}}{{{{\left( {x + 2} \right)}^2}}}} \right].\dfrac{{x + 2}}{4}\\
= \dfrac{{x\left( {x + 2} \right) - {x^2} - 2x - 4}}{{{{\left( {x + 2} \right)}^2}}}.\dfrac{{x + 2}}{4}\\
= \dfrac{{{x^2} + 2x - {x^2} - 2x - 4}}{{{{\left( {x + 2} \right)}^2}}}.\dfrac{{x + 2}}{4}\\
= \dfrac{{ - 4}}{{{{\left( {x + 2} \right)}^2}}}.\dfrac{{x + 2}}{4}\\
= - \dfrac{4}{{x + 2}}\\
b){x^2} - 2x = 0\\
\to x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2\left( l \right)
\end{array} \right.\\
Thay:x = 0\\
\to A = - \dfrac{4}{{0 + 2}} = - 2\\
c)A = 3\\
\to - \dfrac{4}{{x + 2}} = 3\\
\to - 4 = 3x + 6\\
\to 3x = - 10\\
\to x = - \dfrac{{10}}{3}\\
d)A < 1\\
\to - \dfrac{4}{{x + 2}} < 1\\
\to \dfrac{{ - 4 - x - 2}}{{x + 2}} < 0\\
\to \dfrac{{ - 6 - x}}{{x + 2}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- 6 - x > 0\\
x + 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
- 6 - x < 0\\
x + 2 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- 6 > x\\
x < - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x > - 6\\
x > - 2
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > - 2\\
x < - 6
\end{array} \right.\\
e)A.x = - \dfrac{4}{{x + 2}}.x\\
= - \dfrac{{4x}}{{x + 2}} = - \dfrac{{4\left( {x + 2} \right) - 8}}{{x + 2}}\\
= - 4 + \dfrac{8}{{x + 2}}\\
Ax \in Z \to \dfrac{8}{{x + 2}} \in Z\\
\to x + 2 \in U\left( 8 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 8\\
x + 2 = - 8\\
x + 2 = 4\\
x + 2 = - 4\\
x + 2 = 2\\
x + 2 = - 2\\
x + 2 = 1\\
x + 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x = - 10\\
x = 2\left( l \right)\\
x = - 6\\
x = 0\\
x = - 4\\
x = - 1\\
x = - 3
\end{array} \right.
\end{array}\)
