Giai hpt $\left \{ {{3x^{2}+2y^{2}-4xy=11- \frac{1}{y}(2x+\frac{1}{y}) } \atop {2x+\frac{1}{y}-y=4}} \right.$ 10/07/2021 Bởi Mary Giai hpt $\left \{ {{3x^{2}+2y^{2}-4xy=11- \frac{1}{y}(2x+\frac{1}{y}) } \atop {2x+\frac{1}{y}-y=4}} \right.$
$\begin{array}{l} \left\{ \begin{array}{l} 3{x^2} + 2{y^2} – 4xy = 11 – \dfrac{1}{y}\left( {2x + \dfrac{1}{y}} \right)\left( 1 \right)\\ 2x + \dfrac{1}{y} – y = 4\left( 2 \right) \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \dfrac{1}{y} = y + 4 – 2x\\ 3{x^2} + 2{y^2} – 4xy = 11 – \dfrac{1}{y}\left( {2x + \dfrac{1}{y}} \right) \end{array} \right.\\ 3{x^2} + 2{y^2} – 4xy = 11 – \left( {y + 4 – 2x} \right)\left( {2x + y + 4 – 2x} \right)\\ \Leftrightarrow 3{x^2} + 2{y^2} – 4xy = 11 – \left( {y + 4 – 2x} \right)\left( {y + 4} \right)\\ \Leftrightarrow 3{x^2} + 2{y^2} – 4xy = 11 – \left( {{y^2} + 8y + 16 – 2xy – 8x} \right)\\ \Leftrightarrow 3{x^2} + 2{y^2} – 4xy = – 5 – {y^2} – 8y + 2xy + 8x\\ \Leftrightarrow 3{x^2} + 3{y^2} + 8y – 8x – 6xy + 5 = 0\\ \Leftrightarrow 3{\left( {x – y} \right)^2} – 8\left( {x – y} \right) + 5 = 0\\ \Leftrightarrow \left( {x – y – 1} \right)\left( {3\left( {x – y} \right) – 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = y + 1\\ x = y + \dfrac{5}{3} \end{array} \right.\\ x = y + 1 \Rightarrow 2\left( {y + 1} \right) + \dfrac{1}{y} – y = 4\\ \Leftrightarrow y + \dfrac{1}{y} – 2 = 0\\ \Leftrightarrow {y^2} – 2y + 1 = 0\\ \Leftrightarrow {\left( {y – 1} \right)^2} = 0\\ \Leftrightarrow y = 1 \Rightarrow x = 2\\ + x = y + \dfrac{5}{3} \Rightarrow 2\left( {y + \dfrac{5}{3}} \right) + \dfrac{1}{y} – y = 4\\ \Leftrightarrow y + \dfrac{1}{y} – \dfrac{2}{3} = 0\\ \Leftrightarrow 3{y^2} + 3 – 2y = 0(PTVN)\\ \Rightarrow \left( {x;y} \right) = \left( {2;1} \right) \end{array}$ Bình luận
$\begin{array}{l} \left\{ \begin{array}{l} 3{x^2} + 2{y^2} – 4xy = 11 – \dfrac{1}{y}\left( {2x + \dfrac{1}{y}} \right)\left( 1 \right)\\ 2x + \dfrac{1}{y} – y = 4\left( 2 \right) \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \dfrac{1}{y} = y + 4 – 2x\\ 3{x^2} + 2{y^2} – 4xy = 11 – \dfrac{1}{y}\left( {2x + \dfrac{1}{y}} \right) \end{array} \right.\\ 3{x^2} + 2{y^2} – 4xy = 11 – \left( {y + 4 – 2x} \right)\left( {2x + y + 4 – 2x} \right)\\ \Leftrightarrow 3{x^2} + 2{y^2} – 4xy = 11 – \left( {y + 4 – 2x} \right)\left( {y + 4} \right)\\ \Leftrightarrow 3{x^2} + 2{y^2} – 4xy = 11 – \left( {{y^2} + 8y + 16 – 2xy – 8x} \right)\\ \Leftrightarrow 3{x^2} + 2{y^2} – 4xy = – 5 – {y^2} – 8y + 2xy + 8x\\ \Leftrightarrow 3{x^2} + 3{y^2} + 8y – 8x – 6xy + 5 = 0\\ \Leftrightarrow 3{\left( {x – y} \right)^2} – 8\left( {x – y} \right) + 5 = 0\\ \Leftrightarrow \left( {x – y – 1} \right)\left( {3\left( {x – y} \right) – 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = y + 1\\ x = y + \dfrac{5}{3} \end{array} \right.\\ x = y + 1 \Rightarrow 2\left( {y + 1} \right) + \dfrac{1}{y} – y = 4\\ \Leftrightarrow y + \dfrac{1}{y} – 2 = 0\\ \Leftrightarrow {y^2} – 2y + 1 = 0\\ \Leftrightarrow {\left( {y – 1} \right)^2} = 0\\ \Leftrightarrow y = 1 \Rightarrow x = 2\\ + x = y + \dfrac{5}{3} \Rightarrow 2\left( {y + \dfrac{5}{3}} \right) + \dfrac{1}{y} – y = 4\\ \Leftrightarrow y + \dfrac{1}{y} – \dfrac{2}{3} = 0\\ \Leftrightarrow 3{y^2} + 3 – 2y = 0(PTVN)\\ \Rightarrow \left( {x;y} \right) = \left( {2;1} \right) \end{array}$