Giải giúp tớ hệ pt này với x+y+x ² +y ²=8 xy(x+1)(y+1)=12

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1. \[\begin{array}{l}
   \left\{ \begin{array}{l}
   x + y + {x^2} + {y^2} = 8\\
   xy\left( {x + 1} \right)\left( {y + 1} \right) = 12
   \end{array} \right.\\
   \Leftrightarrow \left\{ \begin{array}{l}
   x\left( {x + 1} \right) + y\left( {y + 1} \right) = 8\\
   x\left( {x + 1} \right).y\left( {y + 1} \right) = 12
   \end{array} \right.\,\,\,\left( I \right)\\
   Dat\,\,\left\{ \begin{array}{l}
   x\left( {x + 1} \right) = u\\
   y\left( {y + 1} \right) = v
   \end{array} \right.\,\,\,\left( {{u^2} \ge 4v} \right)\\
   \Rightarrow \left( I \right) \Leftrightarrow \left\{ \begin{array}{l}
   u + v = 8\\
   uv = 12
   \end{array} \right.\\
   \Rightarrow u,\,\,v\,\,\,la\,\,\,2\,\,nghiem\,\,\,cua\,\,pt:\,\,\,{t^2} – 8t + 12 = 0\\
   \Leftrightarrow \left[ \begin{array}{l}
   t = 6\\
   t = 2
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   u = 6\\
   v = 2
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   u = 2\\
   v = 6
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x\left( {x + 1} \right) = 6\\
   y\left( {y + 1} \right) = 2
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x\left( {x + 1} \right) = 2\\
   y\left( {y + 1} \right) = 6
   \end{array} \right.
   \end{array} \right.\\
   \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   {x^2} + x – 6 = 0\\
   {y^2} + y – 2 = 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   {x^2} + x – 2 = 0\\
   {y^2} + y – 6 = 0
   \end{array} \right.
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   \left[ \begin{array}{l}
   x = 2\\
   x = – 3
   \end{array} \right.\\
   \left[ \begin{array}{l}
   y = 1\\
   y = – 2
   \end{array} \right.
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   \left[ \begin{array}{l}
   x = 1\\
   x = – 2
   \end{array} \right.\\
   \left[ \begin{array}{l}
   y = 2\\
   y = – 3
   \end{array} \right.
   \end{array} \right.
   \end{array} \right.
   \end{array}\]
   Em kết luận nghiệm nhé.

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2. $$\eqalign{
   & \left\{ \matrix{
   x + y + {x^2} + {y^2} = 8 \hfill \cr
   xy\left( {x + 1} \right)\left( {y + 1} \right) = 12 \hfill \cr} \right. \cr
   & \Leftrightarrow \left\{ \matrix{
   x\left( {x + 1} \right) + y\left( {y + 1} \right) = 8 \hfill \cr
   x\left( {x + 1} \right)y\left( {y + 1} \right) = 12 \hfill \cr} \right. \cr
   & \Rightarrow x\left( {x + 1} \right)\,\,va\,\,y\left( {y + 1} \right)\,\,la\,\,nghiem\,\,cua\,\,pt \cr
   & {X^2} – 8X + 12 = 0 \Leftrightarrow \left[ \matrix{
   X = 6 \hfill \cr
   X = 2 \hfill \cr} \right. \cr
   & TH1:\,\,\left\{ \matrix{
   x\left( {x + 1} \right) = 6 \hfill \cr
   y\left( {y + 1} \right) = 2 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
   \left[ \matrix{
   x = 2 \hfill \cr
   x = – 3 \hfill \cr} \right. \hfill \cr
   \left[ \matrix{
   y = 1 \hfill \cr
   y = – 2 \hfill \cr} \right. \hfill \cr} \right. \cr
   & TH2:\,\,\left\{ \matrix{
   x\left( {x + 1} \right) = 2 \hfill \cr
   y\left( {y + 1} \right) = 6 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
   \left[ \matrix{
   x = 1 \hfill \cr
   x = – 2 \hfill \cr} \right. \hfill \cr
   \left[ \matrix{
   y = 2 \hfill \cr
   y = – 3 \hfill \cr} \right. \hfill \cr} \right. \cr
   & \Rightarrow S = \left\{ {\left( {2;1} \right);\left( {2; – 2} \right);\left( { – 3;1} \right);\left( { – 3; – 2} \right);\left( {1;2} \right);\left( {1; – 3} \right);\left( { – 2;2} \right);\left( { – 2; – 3} \right)} \right\} \cr} $$

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