Đáp án:
\(\begin{array}{l}
a)\,\,DKXD:\,\,x \ne \pm 2;\,\,x \ne 0.\\
A = \frac{{\left( {7{x^2} + 4} \right)x}}{{\left( {x + 2} \right)\left( {x - 3} \right)}}\\
b)\,\,\,A > 0 \Leftrightarrow \,x > 3\,\,hoac\,\, - 2 < x < 0\\
c)\,\,A = \frac{{9361}}{{104}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \left( {\frac{{2 + x}}{{2 - x}} - \frac{{4{x^2}}}{{{x^2} - 4}} - \frac{{2x}}{{2 + x}}} \right):\left( {\frac{{{x^2} - 3x}}{{2{x^2} - {x^3}}}} \right)\\
a)\,\,DKXD:\,\,x \ne \pm 2;\,\,x \ne 0.\\
A = \left( {\frac{{2 + x}}{{2 - x}} - \frac{{4{x^2}}}{{{x^2} - 4}} - \frac{{2x}}{{2 + x}}} \right):\left( {\frac{{{x^2} - 3x}}{{2{x^2} - {x^3}}}} \right)\\
A = \frac{{ - {{\left( {2 + x} \right)}^2} - 4{x^2} - 2x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\frac{{x\left( {x - 3} \right)}}{{{x^2}\left( {2 - x} \right)}}\\
A = \frac{{ - {x^2} - 4x - 4 - 4{x^2} - 2{x^2} + 4x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\frac{{x\left( {2 - x} \right)}}{{x - 3}}\\
A = \frac{{ - 7{x^2} - 4}}{{x + 2}}.\frac{{ - x}}{{x - 3}}\\
A = \frac{{\left( {7{x^2} + 4} \right)x}}{{\left( {x + 2} \right)\left( {x - 3} \right)}}\\
b)\,\,\,A > 0 \Leftrightarrow \frac{{\left( {7{x^2} + 4} \right)x}}{{\left( {x + 2} \right)\left( {x - 3} \right)}} > 0\\
Do\,\,7{x^2} + 4 > 0\\
\Rightarrow \frac{x}{{\left( {x + 2} \right)\left( {x - 3} \right)}} > 0\\
TH1:\,\,x > 0;\,\,\left( {x + 2} \right)\left( {x - 3} \right) > 0\\
\left( {x + 2} \right)\left( {x - 3} \right) > 0 \Rightarrow x > 3\,\,hoac\,\,x < - 2\\
\Rightarrow x > 3\\
Th2:\,\,x < 0;\,\,\left( {x + 2} \right)\left( {x - 3} \right) < 0\\
\Rightarrow - 2 < x < 3\\
\Rightarrow - 2 < x < 0\\
Vay\,\,x > 3\,\,hoac\,\, - 2 < x < 0\\
c)\,\,\left| {x - 7} \right| = 4 \Leftrightarrow \left[ \begin{array}{l}
x - 7 = 4\\
x - 7 = - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 11\\
x = 3
\end{array} \right.\\
Thay\,\,x = 11 \Rightarrow A = \frac{{\left( {{{7.11}^2} + 4} \right).11}}{{\left( {11 + 2} \right)\left( {11 - 3} \right)}} = \frac{{9361}}{{104}}\\
Thay\,\,x = 3 \Rightarrow A = \frac{{\left( {{{7.3}^2} + 4} \right).11}}{{\left( {3 + 2} \right)\left( {3 - 3} \right)}}\,\,khong\,\,xac\,\,dinh
\end{array}\)
