Đáp án:
a) \(\dfrac{{4x}}{{x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {2 - x} \right)}^2} + 4{x^2}}}{{\left( {2 - x} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {2 - x} \right)\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{{x^2} + 4x + 4 - 4 + 4x - {x^2} + 4{x^2}}}{{\left( {2 - x} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {2 - x} \right)\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{4{x^2} + 8x}}{{\left( {2 - x} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {2 - x} \right)\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{4x\left( {x + 2} \right)}}{{\left( {2 - x} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {2 - x} \right)\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{4x}}{{x - 3}}\\
b)\left| x \right| = \dfrac{1}{3}\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = - \dfrac{1}{3}
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = - \dfrac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \dfrac{{4.\dfrac{1}{3}}}{{\dfrac{1}{3} - 3}} = - \dfrac{1}{2}\\
A = \dfrac{{4.\left( { - \dfrac{1}{3}} \right)}}{{ - \dfrac{1}{3} - 3}} = \dfrac{2}{5}
\end{array} \right.\\
c)A \le 1\\
\to \dfrac{{4x}}{{x - 3}} \le 1\\
\to \dfrac{{4x - x + 3}}{{x - 3}} \le 0\\
\to \dfrac{{3x + 3}}{{x - 3}} \le 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x + 3 \ge 0\\
x - 3 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + 3 \le 0\\
x - 3 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
3 > x \ge - 1\\
\left\{ \begin{array}{l}
x \le - 1\\
x > 3
\end{array} \right.\left( l \right)
\end{array} \right.\\
d)A = \dfrac{{4x}}{{x - 3}} = \dfrac{{4\left( {x - 3} \right) + 12}}{{x - 3}}\\
= 4 + \dfrac{{12}}{{x - 3}}\\
A \in Z \to \dfrac{{12}}{{x - 3}} \in Z\\
\to x - 3 \in U\left( {12} \right)\\
\to \left[ \begin{array}{l}
x - 3 = 12\\
x - 3 = - 12\\
x - 3 = 6\\
x - 3 = - 6\\
x - 3 = 4\\
x - 3 = - 4\\
x - 3 = 3\\
x - 3 = - 3\\
x - 3 = 2\\
x - 3 = - 2\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 15\\
x = - 9\left( l \right)\\
x = 9\\
x = - 3\left( l \right)\\
x = 7\\
x = - 1\left( l \right)\\
x = 6\\
x = 0\left( l \right)\\
x = 5\\
x = 1\left( l \right)\\
x = 4\left( l \right)\\
x = 2\left( l \right)
\end{array} \right.
\end{array}\)
