Да се реши системата

[ammornil]

Задачата е от Нациоална математическа олимпиада на Австрия (1997 г.)$\\[12pt] \begin{array}{|l} (x-1)(y^{2}+6)= y(x^{2}+1) \\[6pt] (y-1)(x^{2}+6)= x(y^{2}+1) \end{array}\\[24pt]$

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Отг. $(2,2), \quad (2,3), \quad (3,2), \quad (3,3)$

[Darina73]

[tex]\begin{array}{|l} x y^{2 }+6x- y^{2 }-6= x^{2 }y+y \\ x^{2 }y+6y- x^{2 } -6=xy^{2 } +x\end{array}[/tex]
Събираме и изваждаме двете уравнения .
[tex]\begin{array}{|l} 6x- y^{2 } -6+6y- x^{2 }-6 = y+x |.(-1) \ne0 \\ 2x y^{2 }+6x- y^{2 }-6-2 x^{2 }y-6y+ x^{2 }+6 = y-x |.(-1) \ne0 \end{array}[/tex]

[tex]\begin{array}{|l} x^{2 } +y^{2 } -5x -5 y +12=0 \\ 2xy(x - y )-7(x-y)-(x-y)(x+y)= 0 \end{array}[/tex]

[tex]\begin{array}{|l}( x+y)^{2 } -2xy-5(x + y )+12=0\\ (x - y )(2xy-7-x-y)=0 \end{array}[/tex]

[tex]\begin{array}{|l} (x+y)^{2 }-2xy-5( x + y)+12 = 0 (А)\\ x = y , или 2xy= (x+y)+7 \end{array}[/tex]

1 случай x=y ,тогава [tex]y^{3 } +6y-y^{2 } -6=y^{3 } +y[/tex]
[tex]y^{2 }-5y+6=0 ; y_{1 } =2[/tex] ,или [tex]y_{2 }[/tex]=3
Получаваме два подслучая : [tex]y_{1 }[/tex]=2 и [tex]x_{1 }[/tex]=2 ,или [tex]y_{2 }[/tex]=3 и [tex]x_{2 }[/tex]=3

2 случай 2xy=(x+y)+7 (заместваме в (А))
[tex](x+y)^{2 }[/tex]-6(x+y)+5=0 x+y= ?
k= -3 ;D=9-5=4 ;[tex](x+y)_{1,2 }= \frac{3 \pm2 }{1}[/tex] т.е. x+y=1 ,или x+y=5
Получаваме два подслучая :
[tex]x_{3 }[/tex] =1-y и [tex]x_{4 }[/tex]= 5-y (Заместваме ги в първото дадено ур-е.)
-[tex]y_{3 }( y_{3 } ^{2 }+6) = y_{3 }(2-2 y_{3 } + y_{3 } ^{2 })[/tex]
Тогава [tex]y_{3 }[/tex]= 0 , или [tex]y_{3 } ^{2 }- y_{3 } +4=0[/tex]
при които [tex]x_{3 }[/tex]=1 ,или няма реални корени
При проверка се вижда ,че (1 ;0) не е корен на второто дадено уравнение и отпада .

[tex]x_{4 }[/tex]=5-y (B) [tex]\Rightarrow[/tex] (4-y)([tex]y^{2 }+6) =y(26-10y+ y^{2 })[/tex]
Получаваме [tex]y_{4 } ^{3 } -7 y_{4 } ^{2 }+16 y_{4 }-12= 0[/tex]
Чрез метода на Хорнер получаваме [tex]y_{4 }[/tex]=2 ,или [tex]y_{4 }[/tex]=3 Връщаме се към (В)
[tex]x_{4 }[/tex]=3 ,или [tex]x_{4 }[/tex]=2
_____________________________________________
При 1 случай получаваме отговори (2;2) и (3;3)
При 2 случай получаваме отговори (3;2) и (2;3)

[ammornil]

Ето още едно предложение за решение.$\\[12pt] \begin{array}{|l} (x-1)(y^{2}+6)= y(x^{2}+1) \\[6pt] (y-1)(x^{2}+6)= x(y^{2}+1) \end{array} \\[24pt] \begin{array}{|l} xy^{2} +6x -y^{2} -6= yx^{2} +y \\[6pt] yx^{2} +6y -x^{2} -6= xy^{2} +x \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} xy^{2} +6x -y^{2} -6- yx^{2} -y= 0 \quad (1) \\[6pt] yx^{2} +6y -x^{2} -6 -xy^{2} -x=0 \quad (2) \end{array} \\[12pt] (1) +(2):\quad \cancel{xy^{2}} +6x -y^{2} -6 \cancel{-yx^{2}} -y \cancel{+ yx^{2}} +6y -x^{2} -6 \cancel{-xy^{2}} -x= 0 \quad \Leftrightarrow \quad \\[6pt] \hspace{5em} -(y^{2} -5y +x^{2} -5x +12)=0 \quad |\cdot{(-1)} \Leftrightarrow \quad \\[6pt] \hspace{5em} y^{2}-2\cdot{\dfrac{5}{2}}\cdot{y} +\left(\dfrac{5}{2}\right)^{2} -\left(\dfrac{5}{2}\right)^{2} +x^{2} -2\cdot{\dfrac{5}{2}}\cdot{x} +\left(\dfrac{5}{2}\right)^{2} -\left(\dfrac{5}{2}\right)^{2} +12=0 \quad \Leftrightarrow \quad \\[6pt] \hspace{5em} \left(x-\dfrac{5}{2}\right)^{2} +\left(y-\dfrac{5}{2}\right)^{2}= \dfrac{1}{2} \quad (A) \\[12pt] (1)-(2): \quad xy^{2} +6x -y^{2} \cancel{-6} -yx^{2} -y -yx^{2} -6y +x^{2} \cancel{+6} +xy^{2} +x= 0 \quad \Leftrightarrow \quad \\[6pt] \hspace{5em} -xy(-y +x +x -y) +(x^{2}-y^{2}) +6(x-y) +(x-y)=0 \quad \Leftrightarrow \quad \\[6pt] \hspace{5em} -2xy(x-y)+(x-y)(x+y)+7(x-y)= 0 \quad \Leftrightarrow \quad \\[6pt] \hspace{5em} (x-y)(x +y -2xy +7)=0 \Rightarrow \text{ (Б1) } \quad x-y=0 \quad \cup \quad \text{ (Б2) } \quad x +y -2xy +7= 0 \\[12pt] \text{(случай 1):} \\[6pt] x=y \Rightarrow (A): \quad \left(y-\dfrac{5}{2}\right)^{2} +\left(y-\dfrac{5}{2}\right)^{2}= \dfrac{1}{2} \\[6pt] \hspace{7em} \left(y-\dfrac{5}{2}\right)^{2}= \dfrac{1}{4} \quad \Leftrightarrow \quad y^{2}-5y +\dfrac{25}{4} -\dfrac{1}{4}=0 \quad \Leftrightarrow \quad y^{2} -5y +6= 0 \\[6pt] \hspace{7em} y_{1,2}= \dfrac{5\pm\sqrt{25 -4\cdot{1}\cdot{6}}}{2\cdot{1}}= \dfrac{5\pm{1}}{2} \Rightarrow x_{1}=y_{1}=2, \quad x_{2}=y_{2}=3 \\[12pt] \text{(случай 2):} \\[6pt] x +y -2xy +7= 0 \quad \Leftrightarrow \quad -2\left(xy -\dfrac{1}{2}x +\dfrac{1}{4} -\dfrac{1}{2}y \right) +\dfrac{1}{2} +7=0 \quad \Leftrightarrow \quad \\[6pt] \hspace{7em} -2\left[x\left(y- \dfrac{1}{2}\right)-\dfrac{1}{2}\left(y -\dfrac{1}{2} \right)\right] +\dfrac{15}{2}= 0 \quad |\div{(-2)} \quad \Leftrightarrow \quad \\[6pt] \hspace{7em} \left(x -\dfrac{1}{2} \right)\cdot{}\left(y -\dfrac{1}{2} \right)= \dfrac{15}{4} \\[12pt] (A): \quad \left(x-\dfrac{5}{2}\right)^{2} +\left(y-\dfrac{5}{2}\right)^{2}= \dfrac{1}{2} \\[6pt] \text{ (Б2) } \quad \left(x -\dfrac{1}{2} \right)\cdot{}\left(y -\dfrac{1}{2} \right)= \dfrac{15}{4} \\[6pt] \quad u=x-\dfrac{5}{2}, \quad v= y-\dfrac{5}{2}, \quad -\dfrac{1}{2}= -\dfrac{5}{2} +2 \\[12pt] \begin{array}{|l} u^{2} +v^{2}= \dfrac{1}{2} \\[6pt] (u+2)(v+2) =\dfrac{15}{4} \end{array} \Leftrightarrow \begin{array}{|l} u^{2} +v^{2}= \dfrac{1}{2} \\[6pt] uv +2u +2v +4 -\dfrac{15}{4}= 0 \quad |\cdot{2} \end{array} \Leftrightarrow \begin{array}{|l} u^{2} +v^{2}= \dfrac{1}{2} \\[6pt] 2uv +4(u +v) = -\dfrac{1}{2} \end{array} \\[6pt] \Leftrightarrow \begin{array}{|l} u^{2} +2uv +v^{2} +4(u +v)= 0 \\[6pt] u^{2} -2uv +v^{2} -4(u +v) = 1 \end{array} \Leftrightarrow \begin{array}{|l} (u+v)^{2} +4(u +v)= 0 \\[6pt] (u-v)^{2} -4(u +v) = 1 \end{array} \Rightarrow \\[6pt] \begin{array}{|l} u+v= 0 \\[6pt] (u-v)^{2} -4(u +v) = 1 \end{array} \quad \cup \quad \begin{array}{|l} u +v =-4 \\[6pt] (u-v)^{2} -4(u +v) = 1 \end{array} \\[6pt] \begin{array}{|l} u= -v \\[6pt] 4v^{2} = 1 \end{array} \quad \cup \quad \begin{array}{|l} u= v -4 \\[6pt] 4^{2} +16 = 1 \end{array} \\[6pt] \begin{array}{|l} u= -v \\[6pt] v_{1,2}= \pm{\dfrac{1}{4}} \end{array} \quad \cup \quad \oslash \\[12pt] u_{1}= \dfrac{1}{2}, v_{1}= - \dfrac{1}{2} \Rightarrow x_{1}= 3, y_{1}= 2 \\[6pt] u_{2}= -\dfrac{1}{2}, v_{2}= \dfrac{1}{2} \Rightarrow x_{2}= 2, y_{2}= 3 \\[24pt]$ $$ \text{Отг.}\quad (2,2), \quad (3,3), \quad (2,3) \quad (3,2)$$

[ammornil]

В горното съм допуснал грешка, която не променя отговора но все пак е грешка ... $\\[6pt] \Leftrightarrow \begin{array}{|l} u^{2} +2uv +v^{2} +4(u +v)= 0 \\[6pt] u^{2} -2uv +v^{2} -4(u +v) = 1 \end{array} \Leftrightarrow \begin{array}{|l} (u+v)^{2} +4(u +v)= 0 \\[6pt] (u-v)^{2} -4(u +v) = 1 \end{array} \Rightarrow \\[6pt] \begin{array}{|l} u+v= 0 \\[6pt] (u-v)^{2} -4(u +v) = 1 \end{array} \quad \cup \quad \begin{array}{|l} u +v =-4 \\[6pt] (u-v)^{2} -4(u +v) = 1 \end{array} \\[6pt] \begin{array}{|l} u= -v \\[6pt] 4v^{2} = 1 \end{array} \quad \cup \quad \begin{array}{|l} \red{u= -v -4} \\[6pt] \red{4v^{2} +16v +16 +16 - 1= 0 } \end{array} \\[6pt] \begin{array}{|l} u= -v \\[6pt] v_{1,2}= \pm{\dfrac{1}{4}} \end{array} \quad \cup \quad \begin{array}{|l} u=-v-4 \\ 4v^{2} +16v +255 =0 \begin{cases} a>0 \\ D<0 \end{cases} \end{array} \Rightarrow \oslash \\[12pt] u_{1}= \dfrac{1}{2}, v_{1}= - \dfrac{1}{2} \Rightarrow x_{1}= 3, y_{1}= 2 \\[6pt] u_{2}= -\dfrac{1}{2}, v_{2}= \dfrac{1}{2} \Rightarrow x_{2}= 2, y_{2}= 3 \\[24pt]$ $$ \text{Отг.}\quad (2,2), \quad (3,3), \quad (2,3) \quad (3,2)\\[12pt]$$

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$(-2v-4)^{2}=[-1\cdot(2v+4)]^{2}=(-1)^{2}\cdot{(2v+4)^{2}}= (2v)^{2}+2\cdot{2v}\cdot{4}+4^{2}= 4v^{2} +16v +16$