giải hpt : 1) x=2y/1-y^2 và y=2x/1-x^2 2) x^3=5x+y và y^3=5y+x

Đáp án:

1) $S = \left\{ {\left( {0;0} \right),\left( {\sqrt {15} ;\dfrac{{\sqrt {15} }}{5}} \right),\left( { - \sqrt {15} ;\dfrac{{ - \sqrt {15} }}{5}} \right)} \right\}$

2) $S = \left\{ {\left( {0;0} \right),\left( {\sqrt 6 ;\sqrt 6 } \right),\left( { - \sqrt 6 ; - \sqrt 6 } \right),\left( {2; - 2} \right),\left( { - 2;2} \right),\left( {\dfrac{{\sqrt 7  - \sqrt 3 }}{2};\dfrac{{ - \sqrt 7  - \sqrt 3 }}{2}} \right),\left( {\dfrac{{ - \sqrt 7  - \sqrt 3 }}{2};\dfrac{{\sqrt 7  - \sqrt 3 }}{2}} \right),\left( {\dfrac{{\sqrt 7  + \sqrt 3 }}{2};\dfrac{{ - \sqrt 7  + \sqrt 3 }}{2}} \right),\left( {\dfrac{{ - \sqrt 7  + \sqrt 3 }}{2};\dfrac{{\sqrt 7  + \sqrt 3 }}{2}} \right)} \right\}$

Giải thích các bước giải:

 1) ĐKXĐ: $x,y\ne \pm 1$

Ta có:

$\begin{array}{l} \left\{ \begin{array}{l} x = \dfrac{{2y}}{{1 - {y^2}}}\\ y = \dfrac{{2x}}{{1 - {x^2}}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x\left( {1 - {y^2}} \right) = 2y\\ y\left( {1 - {x^2}} \right) = 2x \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} x\left( {1 - {y^2}} \right) = 2y\\ x\left( {1 - {y^2}} \right) - y\left( {1 - {x^2}} \right) = 2\left( {y - x} \right) \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} x\left( {1 - {y^2}} \right) = 2y\\ x - y - x{y^2} + {x^2}y + 2\left( {x - y} \right) = 0 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} x\left( {1 - {y^2}} \right) = 2y\\ \left( {x - y} \right)\left( {3 - xy} \right) = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x\left( {1 - {y^2}} \right) = 2y\left( 1 \right)\\ \left[ \begin{array}{l} x = y\\ xy = 3 \end{array} \right. \end{array} \right.\\  + )TH1:x = y\\ \left( 1 \right)tt:y\left( {1 - {y^2}} \right) = 2y\\  \Leftrightarrow y\left( {{y^2} + 1} \right) = 1\\  \Leftrightarrow y = 0 \Rightarrow x = y = 0\left( {tm} \right)\\  + )TH2:xy = 3 \Rightarrow x,y \ne 0\\ \left( 1 \right) \Leftrightarrow x - x{y^2} = 2y\\  \Leftrightarrow x - xy.y = 2y\\  \Leftrightarrow x - 3y = 2y\\  \Leftrightarrow x = 5y\\  \Leftrightarrow \dfrac{3}{y} = 5y\\  \Leftrightarrow {y^2} = \dfrac{3}{5}\\  \Leftrightarrow \left[ \begin{array}{l} y = \dfrac{{\sqrt {15} }}{5}\\ y = \dfrac{{ - \sqrt {15} }}{5} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \sqrt {15} ;y = \dfrac{{\sqrt {15} }}{5}\\ x =  - \sqrt {15} ;y = \dfrac{{ - \sqrt {15} }}{5} \end{array} \right.\left( {tm} \right)\\ \end{array}$

Vậy hệ có tập nghiệm: $S = \left\{ {\left( {0;0} \right),\left( {\sqrt {15} ;\dfrac{{\sqrt {15} }}{5}} \right),\left( { - \sqrt {15} ;\dfrac{{ - \sqrt {15} }}{5}} \right)} \right\}$

2) Ta có:

$\begin{array}{l} \left\{ \begin{array}{l} {x^3} = 5x + y\\ {y^3} = 5y + x \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ {x^3} - {y^3} = 5x + y - \left( {5y + x} \right) \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ {x^3} - {y^3} - 4\left( {x - y} \right) = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ \left( {x - y} \right)\left( {{x^2} + xy + {y^2} - 4} \right) = 0 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\left( 1 \right)\\ \left[ \begin{array}{l} x - y = 0\\ {x^2} + xy + {y^2} - 4 = 0 \end{array} \right. \end{array} \right.\\  + )TH1:x - y = 0 \Leftrightarrow x = y\\ \left( 1 \right)tt:{x^3} = 6x\\  \Leftrightarrow {x^3} - 6x = 0\\  \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \sqrt 6 \\ x =  - \sqrt 6  \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = y = 0\\ x = y = \sqrt 6 \\ x = y =  - \sqrt 6  \end{array} \right.\\  + )TH2:{x^2} + xy + {y^2} - 4 = 0\\ \left\{ \begin{array}{l} {x^3} = 5x + y\\ {x^2} + xy + {y^2} - 4 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ {x^2} + xy + {y^2} = 4 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ 4{x^3} = \left( {5x + y} \right)\left( {{x^2} + xy + {y^2}} \right) \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ {x^3} + {y^3} + 6{x^2}y + 6x{y^2} = 0 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ \left( {x + y} \right)\left( {{x^2} + 5xy + {y^2}} \right) = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^3} = 5x + y\\ \left[ \begin{array}{l} y =  - x\\ {x^2} + 5xy + {y^2} = 0 \end{array} \right. \end{array} \right.\\ *)y =  - x\\ \left( 1 \right)tt:{x^3} = 5x - x\\  \Leftrightarrow {x^3} - 4x = 0\\  \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 2\\ x =  - 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = y = 0\\ x = 2;y =  - 2\\ x =  - 2;y = 2 \end{array} \right.\\ *){x^2} + 5xy + {y^2} = 0\\ \left\{ \begin{array}{l} {x^2} + 5xy + {y^2} = 0\\ {x^2} + xy + {y^2} = 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} xy =  - 1\\ {x^2} + {y^2} = 5 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} xy =  - 1\\ {\left( {x + y} \right)^2} - 2xy = 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} xy =  - 1\\ {\left( {x + y} \right)^2} = 3 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} xy =  - 1\\ \left[ \begin{array}{l} x + y = \sqrt 3 \\ x + y =  - \sqrt 3  \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x + y = \sqrt 3 \\ xy =  - 1 \end{array} \right.\left( I \right)\\ \left\{ \begin{array}{l} x + y =  - \sqrt 3 \\ xy =  - 1 \end{array} \right.\left( {II} \right) \end{array} \right.\\ \left( I \right) \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{\sqrt 7  + \sqrt 3 }}{2};y = \dfrac{{ - \sqrt 7  + \sqrt 3 }}{2}\\ x = \dfrac{{ - \sqrt 7  + \sqrt 3 }}{2};y = \dfrac{{\sqrt 7  + \sqrt 3 }}{2} \end{array} \right.\\ \left( {II} \right) \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{\sqrt 7  - \sqrt 3 }}{2};y = \dfrac{{ - \sqrt 7  - \sqrt 3 }}{2}\\ x = \dfrac{{ - \sqrt 7  - \sqrt 3 }}{2};y = \dfrac{{\sqrt 7  - \sqrt 3 }}{2} \end{array} \right. \end{array}$

Vậy hệ có tập nghiệm là:

$S = \left\{ {\left( {0;0} \right),\left( {\sqrt 6 ;\sqrt 6 } \right),\left( { - \sqrt 6 ; - \sqrt 6 } \right),\left( {2; - 2} \right),\left( { - 2;2} \right),\left( {\dfrac{{\sqrt 7  - \sqrt 3 }}{2};\dfrac{{ - \sqrt 7  - \sqrt 3 }}{2}} \right),\left( {\dfrac{{ - \sqrt 7  - \sqrt 3 }}{2};\dfrac{{\sqrt 7  - \sqrt 3 }}{2}} \right),\left( {\dfrac{{\sqrt 7  + \sqrt 3 }}{2};\dfrac{{ - \sqrt 7  + \sqrt 3 }}{2}} \right),\left( {\dfrac{{ - \sqrt 7  + \sqrt 3 }}{2};\dfrac{{\sqrt 7  + \sqrt 3 }}{2}} \right)} \right\}$