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