giải hpt x^3+y^3=5x+y x^2-y^2=3

Đáp án:

\(S = \left\{ {\left( { - 2;1} \right);\,\,\left( {2; - 1} \right)} \right\}\)

Giải thích các bước giải:

$$\eqalign{ & \left\{ \matrix{ {x^3} + {y^3} = 5x + y \hfill \cr {x^2} - {y^2} = 3\,\,\left( * \right) \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) = 5x + y\,\,\,\left( 1 \right) \hfill \cr \left( {x - y} \right)\left( {x + y} \right) = 3\,\,\,\left( 2 \right) \hfill \cr} \right. \cr & \left( 1 \right):\left( 2 \right) \cr & \Rightarrow {{{x^2} + xy + {y^2}} \over {x + y}} = {{5x + y} \over 3} \cr & \Leftrightarrow 3{x^2} + 3xy + 3{y^2} = 5{x^2} + xy + 5xy + {y^2} \cr & \Leftrightarrow - 2{x^2} - 3xy + 2{y^2} = 0 \cr & \Leftrightarrow \left[ \matrix{ x = {1 \over 2}y \hfill \cr x = - 2y \hfill \cr} \right. \cr & TH1:\,\,x = {1 \over 2}y \Leftrightarrow y = 2x \cr & Thay\,\,\left( * \right):\,\,{x^2} - {\left( {2x} \right)^2} = 3 \Leftrightarrow - 3{x^2} = 3 \Leftrightarrow {x^2} = - 1\,\,\left( {Vo\,\,nghiem} \right) \cr & TH2:\,\,x = - 2y \cr & Thay\,\left( * \right):\,\,{\left( { - 2y} \right)^2} - {y^2} = 3 \cr & \Leftrightarrow 3{y^2} = 3 \Leftrightarrow \left[ \matrix{ y = 1 \Rightarrow x = - 2 \hfill \cr y = - 1 \Rightarrow x = 2 \hfill \cr} \right. \cr & Vay\,\,S = \left\{ {\left( { - 2;1} \right);\,\,\left( {2; - 1} \right)} \right\} \cr} $$