## Giải giùm mìn với mn ơi

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Giải giùm mìn với mn ơi

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1 year 2020-11-22T09:26:42+00:00 1 Answers 30 views 0

$$\begin{array}{l} a,\\ DKXD:\,\,\,\left\{ \begin{array}{l} x \ne 0\\ x \ne \pm 2\\ x \ne 3 \end{array} \right.\\ A = \left( {\dfrac{{2 + x}}{{2 – x}} – \dfrac{{4{x^2}}}{{{x^2} – 4}} – \dfrac{{2 – x}}{{2 + x}}} \right):\left( {\dfrac{{{x^2} – 3x}}{{2{x^2} – {x^3}}}} \right)\\ = \left( { – \dfrac{{x + 2}}{{x – 2}} – \dfrac{{4{x^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} + \dfrac{{x – 2}}{{x + 2}}} \right):\dfrac{{x\left( {x – 3} \right)}}{{{x^2}\left( {2 – x} \right)}}\\ = \dfrac{{ – {{\left( {x + 2} \right)}^2} – 4{x^2} + {{\left( {x – 2} \right)}^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}:\dfrac{{x – 3}}{{x\left( {2 – x} \right)}}\\ = \dfrac{{ – \left( {{x^2} + 4x + 4} \right) – 4{x^2} + \left( {{x^2} – 4x + 4} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}:\dfrac{{x – 3}}{{x\left( {2 – x} \right)}}\\ = \dfrac{{ – 4{x^2} – 8x}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}.\dfrac{{x\left( {2 – x} \right)}}{{x – 3}}\\ = \dfrac{{ – 4x\left( {x + 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}.\dfrac{{x\left( {2 – x} \right)}}{{x – 3}}\\ = \dfrac{{4{x^2}}}{{x – 3}}\\ b,\\ A > 0 \Leftrightarrow \dfrac{{4{x^2}}}{{x – 3}} > 0 \Leftrightarrow x – 3 > 0 \Leftrightarrow x > 3\\ c,\\ \left| {x – 7} \right| = 4 \Leftrightarrow \left[ \begin{array}{l} x – 7 = 4\\ x – 7 = – 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 11\\ x = 3 \end{array} \right.\\ x \ne 3 \Rightarrow x = 11\\ A = \dfrac{{4{x^2}}}{{x – 3}} = \dfrac{{{{4.11}^2}}}{{11 – 3}} = \dfrac{{121}}{2} \end{array}$$