giải HPT $\left \{ {{x(x^2-2)+x^2y+4=2(x^2+y)} \atop {x^2-y+2=0}} \right.$ 20/07/2021 Bởi Parker giải HPT $\left \{ {{x(x^2-2)+x^2y+4=2(x^2+y)} \atop {x^2-y+2=0}} \right.$
$\left \{ {{x(x²-2)+x²y + 4 =2(x²+y)(1)} \atop {x²-y+2=0}} \right.$ Ta xét PT (1) $(1) ⇔ x(x² – 2) + x²y + 4 – 2x² – 2y = 0$ $⇔ x(x² – 2) + x²y – 2y – 2x² + 4 = 0$ $⇔ x(x² – 2) + y(x² – 2) – 2(x² – 2) = 0$ $⇔ (x² – 2)(x + y – 2) = 0$ $⇔\left[ \begin{array}{l}x²-2=0\\x+y-2=0\end{array} \right.$ TH1: $x²-2=0$ , ta có HPT: $\left \{ {{x²-2=0} \atop {x²-y+2=0}} \right.$ $⇔ \left \{ {{x²-2=0} \atop {y-4=0}} \right.$ $⇔ \left \{ {{x²=2} \atop {y=4}} \right.$ $⇔\left[ \begin{array}{l}\left \{ {{x=\sqrt{2}} \atop {y=4}} \right.\\\left \{ {{x=-\sqrt{2}} \atop {y=4}} \right.\end{array} \right.$ TH2: $x + y – 2 = 0$, ta có HPT: $\left \{ {{x + y – 2 = 0} \atop {x²-y+2=0}} \right.$ $⇔ \left \{ {{x + y – 2 = 0} \atop {x+x²=0}} \right.$ $⇔ \left \{ {{x + y – 2 = 0} \atop {x(x + 1) = 0}} \right.$ $⇔\left[ \begin{array}{l}\left \{ {{x=0} \atop {x + y – 2 = 0}} \right.\\\left \{ {{x+1=0} \atop {x + y – 2 = 0}} \right.\end{array} \right.$ $⇔\left[ \begin{array}{l}\left \{ {{x=0} \atop {y=2}} \right.\\\left \{ {{x=-1} \atop {y=3}} \right.\end{array} \right.$ Vậy các cặp nghiệm $(x,y)$ thỏa mãn là: $(\sqrt{2},4)$ , $(-\sqrt{2},4)$ , $(0,2)$ , $(-1,3)$ Bình luận
$\begin{array}{l} \left\{ \begin{array}{l} x\left( {{x^2} – 2} \right) + {x^2}y + 4 = 2\left( {{x^2} + y} \right)\\ {x^2} – y + 2 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x\left( {{x^2} – 2} \right) + {x^2}y + 4 = 2\left( {{x^2} + y} \right)\left( 1 \right)\\ {x^2} = y – 2\left( 2 \right) \end{array} \right.\\ \left( 2 \right) \to \left( 1 \right):x\left( {y – 4} \right) + \left( {y – 2} \right)y + 4 = 2\left( {2y – 2} \right)\\ \Leftrightarrow xy – 4x + {y^2} – 2y + 4 = 4y – 4\\ \Leftrightarrow {y^2} – 6y + 8 + xy – 4x = 0\\ \Leftrightarrow {y^2} + y\left( {x – 6} \right) + 8 – 4x = 0\\ \Leftrightarrow y\left( {x + y – 2} \right) – 4\left( {x + y – 2} \right) = 0\\ \Leftrightarrow \left( {x + y – 2} \right)\left( {y – 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} y = 2 – x\left( 3 \right)\\ y = 4\left( 4 \right) \end{array} \right.\\ \left( 3 \right) \to \left( 2 \right):{x^2} = 4 – 2 \Rightarrow x = \pm \sqrt 2 \\ \left( 4 \right) \to \left( 2 \right):{x^2} – y + 2 = 0 \Leftrightarrow {x^2} – \left( {2 – x} \right) + 2 = 0\\ \Leftrightarrow {x^2} + x = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0 \Rightarrow y = 2\\ x = – 1 \Rightarrow y = 3 \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( { – \sqrt 2 ;4} \right),\left( { – \sqrt 2 ;4} \right),\left( {0;2} \right),\left( { – 1;3} \right) \end{array}$ Bình luận
$\left \{ {{x(x²-2)+x²y + 4 =2(x²+y)(1)} \atop {x²-y+2=0}} \right.$
Ta xét PT (1)
$(1) ⇔ x(x² – 2) + x²y + 4 – 2x² – 2y = 0$
$⇔ x(x² – 2) + x²y – 2y – 2x² + 4 = 0$
$⇔ x(x² – 2) + y(x² – 2) – 2(x² – 2) = 0$
$⇔ (x² – 2)(x + y – 2) = 0$
$⇔\left[ \begin{array}{l}x²-2=0\\x+y-2=0\end{array} \right.$
TH1: $x²-2=0$ , ta có HPT:
$\left \{ {{x²-2=0} \atop {x²-y+2=0}} \right.$
$⇔ \left \{ {{x²-2=0} \atop {y-4=0}} \right.$
$⇔ \left \{ {{x²=2} \atop {y=4}} \right.$
$⇔\left[ \begin{array}{l}\left \{ {{x=\sqrt{2}} \atop {y=4}} \right.\\\left \{ {{x=-\sqrt{2}} \atop {y=4}} \right.\end{array} \right.$
TH2: $x + y – 2 = 0$, ta có HPT:
$\left \{ {{x + y – 2 = 0} \atop {x²-y+2=0}} \right.$
$⇔ \left \{ {{x + y – 2 = 0} \atop {x+x²=0}} \right.$
$⇔ \left \{ {{x + y – 2 = 0} \atop {x(x + 1) = 0}} \right.$
$⇔\left[ \begin{array}{l}\left \{ {{x=0} \atop {x + y – 2 = 0}} \right.\\\left \{ {{x+1=0} \atop {x + y – 2 = 0}} \right.\end{array} \right.$
$⇔\left[ \begin{array}{l}\left \{ {{x=0} \atop {y=2}} \right.\\\left \{ {{x=-1} \atop {y=3}} \right.\end{array} \right.$
Vậy các cặp nghiệm $(x,y)$ thỏa mãn là: $(\sqrt{2},4)$ , $(-\sqrt{2},4)$ , $(0,2)$ , $(-1,3)$
$\begin{array}{l} \left\{ \begin{array}{l} x\left( {{x^2} – 2} \right) + {x^2}y + 4 = 2\left( {{x^2} + y} \right)\\ {x^2} – y + 2 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x\left( {{x^2} – 2} \right) + {x^2}y + 4 = 2\left( {{x^2} + y} \right)\left( 1 \right)\\ {x^2} = y – 2\left( 2 \right) \end{array} \right.\\ \left( 2 \right) \to \left( 1 \right):x\left( {y – 4} \right) + \left( {y – 2} \right)y + 4 = 2\left( {2y – 2} \right)\\ \Leftrightarrow xy – 4x + {y^2} – 2y + 4 = 4y – 4\\ \Leftrightarrow {y^2} – 6y + 8 + xy – 4x = 0\\ \Leftrightarrow {y^2} + y\left( {x – 6} \right) + 8 – 4x = 0\\ \Leftrightarrow y\left( {x + y – 2} \right) – 4\left( {x + y – 2} \right) = 0\\ \Leftrightarrow \left( {x + y – 2} \right)\left( {y – 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} y = 2 – x\left( 3 \right)\\ y = 4\left( 4 \right) \end{array} \right.\\ \left( 3 \right) \to \left( 2 \right):{x^2} = 4 – 2 \Rightarrow x = \pm \sqrt 2 \\ \left( 4 \right) \to \left( 2 \right):{x^2} – y + 2 = 0 \Leftrightarrow {x^2} – \left( {2 – x} \right) + 2 = 0\\ \Leftrightarrow {x^2} + x = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0 \Rightarrow y = 2\\ x = – 1 \Rightarrow y = 3 \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( { – \sqrt 2 ;4} \right),\left( { – \sqrt 2 ;4} \right),\left( {0;2} \right),\left( { – 1;3} \right) \end{array}$