Đáp án:
\(\begin{array}{l}
a)\,\,\,F = \frac{{4 - 2x}}{{x + 2}}.\\
b)\,\,x = - \frac{2}{5}.\\
c)\,\,x < - 2.\\
d)\,\,\,x \in \left\{ { - 10; - 6; - 4; - 3; - 1;\,\,6} \right\}.\\
e)\,\,\,F = \frac{4}{5}.
\end{array}\)
Giải thích các bước giải:
\(F = \left( {\frac{2}{{x + 2}} - \frac{4}{{{x^2} + 4x + 4}}} \right):\left( {\frac{2}{{{x^2} - 4}} + \frac{1}{{2 - x}}} \right)\,\,\,\,\,\left( {x \ne 0;\,\,x \ne \pm 2} \right)\)
a) Rút gọn biểu thức:
\(\begin{array}{l}F = \left( {\frac{2}{{x + 2}} - \frac{4}{{{x^2} + 4x + 4}}} \right):\left( {\frac{2}{{{x^2} - 4}} + \frac{1}{{2 - x}}} \right)\,\,\,\,\,\left( {x \ne 0;\,\,x \ne \pm 2} \right)\\ = \left[ {\frac{2}{{x + 2}} - \frac{4}{{{{\left( {x + 2} \right)}^2}}}} \right]:\left[ {\frac{2}{{\left( {x + 2} \right)\left( {x - 2} \right)}} - \frac{1}{{x - 2}}} \right]\\ = \frac{{2\left( {x + 2} \right) - 4}}{{{{\left( {x + 2} \right)}^2}}}:\frac{{2 - x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\ = \frac{{2x + 4 - 4}}{{{{\left( {x + 2} \right)}^2}}}.\frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{ - x}}\\ = \frac{{2x}}{{x + 2}}.\frac{{x - 2}}{{ - x}} = - \frac{{2\left( {x - 2} \right)}}{{x + 2}} = \frac{{4 - 2x}}{{x + 2}}.\end{array}\)
b) Tìm \(x\) để \(F = 3.\)
Điểu kiện: \(x \ne 0;\,\,x \ne \pm 2.\)
Ta có: \(F = 3\)
\(\begin{array}{l} \Leftrightarrow \frac{{4 - 2x}}{{x + 2}} = 3 \Leftrightarrow 4 - 2x = 3x + 6\\ \Leftrightarrow 5x = - 2 \Leftrightarrow x = - \frac{2}{5}\,\,\,\left( {tm} \right).\end{array}\)
Vậy \(x = - \frac{2}{5}\) thì \(F = 3.\)
c) Tìm \(x\) để \(F < - 2.\)
Điểu kiện: \(x \ne 0;\,\,x \ne \pm 2.\)
\(\begin{array}{l}F < - 2 \Leftrightarrow \frac{{ - 2\left( {x - 2} \right)}}{{x + 2}} < - 2\\ \Leftrightarrow \frac{{x - 2}}{{x + 2}} > 1 \Leftrightarrow \frac{{x - 2}}{{x + 2}} - 1 > 0\\ \Leftrightarrow \frac{{x - 2 - x - 2}}{{x + 2}} > 0\\ \Leftrightarrow \frac{{ - 4}}{{x + 2}} > 0 \Leftrightarrow x + 2 < 0\,\,\,\,\left( {do\,\, - 4 < 0} \right)\\ \Leftrightarrow x < - 2.\end{array}\)
Kết hợp với điều kiện ta được \(x < - 2\) thì \(F < - 2.\)
Vậy \(x < - 2.\)
d) Tìm \(x \in \mathbb{Z}\) để \(F \in \mathbb{Z}.\)
Điểu kiện: \(x \ne 0;\,\,x \ne \pm 2.\)
Ta có: \(F = \frac{{ - 2\left( {x - 2} \right)}}{{x + 2}} = - \frac{{2\left( {x + 2 - 4} \right)}}{{x + 2}} = - \frac{{2\left( {x + 2} \right)}}{{x + 2}} + \frac{8}{{x + 2}} = - 2 + \frac{8}{{x + 2}}.\)
\(\begin{array}{l} \Rightarrow F \in \mathbb{Z} \Leftrightarrow \frac{8}{{x + 2}} \in \mathbb{Z} \Leftrightarrow x + 2 \in U\left( 8 \right) = \left\{ { \pm 1;\,\, \pm 2;\,\, \pm 4;\,\, \pm 8} \right\}\\ \Rightarrow \left[ \begin{array}{l}x + 2 = - 8\\x + 2 = - 4\\x + 2 = - 2\\x + 2 = - 1\\x + 2 = 1\\x + 2 = 2\\x + 2 = 4\\x + 2 = 8\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 10\,\,\,\left( {tm} \right)\\x = - 6\,\,\,\left( {tm} \right)\\x = - 4\,\,\,\left( {tm} \right)\\x = - 3\,\,\,\left( {tm} \right)\\x = - 1\,\,\,\left( {tm} \right)\\x = 0\,\,\,\left( {ktm} \right)\\x = 2\,\,\,\left( {ktm} \right)\\x = 6\,\,\,\left( {tm} \right)\end{array} \right.\end{array}\)
Vậy \(x \in \left\{ { - 10; - 6; - 4; - 3; - 1;\,\,6} \right\}.\)
e) Tìm \(F\) biết \({x^2} - x - 6 = 0.\)
Điểu kiện: \(x \ne 0;\,\,x \ne \pm 2.\)
Ta có: \({x^2} - x - 6 = 0 \Leftrightarrow {x^2} - 3x + 2x - 6 = 0\)
\(\begin{array}{l} \Leftrightarrow x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0 \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 3 = 0\\x + 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\,\,\,\left( {tm} \right)\\x = - 2\,\,\,\left( {ktm} \right)\end{array} \right..\end{array}\)
Với \(x = 3\) ta có: \(F = \frac{{ - 4\left( {2 - x} \right)}}{{x + 2}} = \frac{{ - 4\left( {2 - 3} \right)}}{{3 + 2}} = \frac{4}{5}.\)
Vậy \(F = \frac{4}{5}.\)
