Đáp án:
$\left \{ {{y^2 + y^2 = 5} \atop {3(x+ y) + 4xy + 5 = 0}} \right.$ `↔` $\left \{ {{(x + y)^2 - 2xy = 5} \atop {3(x+ y) + 4xy + 5 = 0}} \right.$ `(1)`
Đặt `x+ y = a , xy = b (a^2 >= 4b)`
`(1) ↔` $\left \{ {{a^2 - 2b = 5} \atop {3a + 4b + 5 = 0}} \right.$ `↔` $\left \{ {{a^2 - 2b = 5} \atop {b = -\dfrac{5 + 3a}{4}}} \right.$ `↔` $\left \{ {{a^2 + 2\dfrac{5 + 3a}{4} = 5} \atop {b = -\dfrac{5 + 3a}{4}}} \right.$
`↔` $\left \{ {{2a^2 + 3a - 5 = 0} \atop {b = -\dfrac{5 + 3a}{4}}} \right.$ `↔` $\left \{ {{(a - 1)(2a + 5)= 0} \atop {b = -\dfrac{5 + 3a}{4}}} \right.$
`↔` $\left \{ {{\left \{ {{a - 1 = 0} \atop {b = -\dfrac{5 + 3a}{4}}} \right.} \atop {\left \{ {{2a + 5 = 0} \atop {b = -\dfrac{5 + 3a}{4}}} \right.}} \right.$ `↔` $\left \{ {{\left \{ {{a = 1} \atop {b = -2}} \right.} \atop {\left \{ {{a = \dfrac{-5}{2}} \atop {b = \dfrac{5}{8}}} \right.}} \right.$
`th1:`
`↔`$\left \{ {{x + y = \dfrac{-5}{2} } \atop {xy = \dfrac{5}{8}}} \right.$ `↔` $\left \{ {{y = \dfrac{-5}{2} - x } \atop {xy = \dfrac{5}{8}}} \right.$ `↔` $\left \{ {{y = \dfrac{-5}{2} - x } \atop {x(\dfrac{-5}{2} - x) = \dfrac{5}{8}}} \right.$
`↔` $\left \{ {{y = \dfrac{-5}{2} - x } \atop {8x^2 + 20x + 5 = 0}} \right.$ `↔` $\left \{ {{\left \{ {{y = \dfrac{-5}{2} - x } \atop {x= \dfrac{\sqrt{15} - 5}{4}}} \right.} \atop {\left \{ {{y=\dfrac{-5}{2} - x } \atop {x = - \dfrac{\sqrt{15} + 5}{4}}} \right.}} \right.$
`↔` $\left \{ {{\left \{ {{y = - \dfrac{\sqrt{15} + 5}{4} } \atop {x= \dfrac{\sqrt{15} - 5}{4}}} \right.} \atop {\left \{ {{y= \dfrac{\sqrt{15} - 5}{4}} \atop {x = - \dfrac{\sqrt{15} + 5}{4}}} \right.}} \right.$
`th2 : `
$\left \{ {{x + y = 1} \atop {xy = -2}} \right.$ `↔` $\left \{ {{y = 1 - x} \atop {xy = -2}} \right.$ `↔` $\left \{ {{y = 1 - x} \atop {x(1 - x) = -2}} \right.$
`↔` $\left \{ {{y = 1 - x} \atop {x^2- x - 2 = 0}} \right.$ `↔` $\left \{ {{y = 1 - x} \atop {(x + 1)(x - 2) = 0}} \right.$
`↔` $\left \{ {{\left \{ {{y=1 - x} \atop {x + 1 = 0}} \right.} \atop {\left \{ {{y=1 - x} \atop {x - 2 = 0}} \right.}} \right.$ `↔` $\left \{ {{\left \{ {{y=2 } \atop {x = -1}} \right.} \atop {\left \{ {{y=-1} \atop {x = 2}} \right.}} \right.$
Vậy `(x,y)` là hoán vị của `(- \frac{\sqrt{15} + 5}{4} ; \frac{\sqrt{15} - 5}{4})` và là hoán vị của `(-1; 2)`
Giải thích các bước giải:
