Решите систему уравнений :
x+y-xy= 1
x^2+y^2-xy=3
И всё это взять в фигурную скобочку.
Помогите, пожалуйста!
Ответы
Ответ дал:
0
Воспользуемся известной формулой:
Подставим в систему:
![begin{cases} (x+y)-xy=1 \ (x+y)^2-2xy-xy=3 end{cases} <=>
begin{cases} (x+y)-xy=1 \ (x+y)^2-3xy=3 end{cases}\\
x+y=a, xy=b\\
begin{cases} a-b=1 \ a^2-3b=3 end{cases}<=> begin{cases} b=a-1 \ a^2-3(a-1)=3 end{cases} <=>\
begin{cases} b=a-1 \ a^2-3a=0 end{cases} <=>
begin{cases} b=a-1 \ a(a-3)=0 end{cases} \
a_1=0 =>b_1=0-1=-1\
a_2=3 =>b_2=3-1=2\](https://tex.z-dn.net/?f=begin%7Bcases%7D+%28x%2By%29-xy%3D1+%5C+%28x%2By%29%5E2-2xy-xy%3D3+end%7Bcases%7D+%26lt%3B%3D%26gt%3B%0Abegin%7Bcases%7D+%28x%2By%29-xy%3D1+%5C+%28x%2By%29%5E2-3xy%3D3+end%7Bcases%7D%5C%5C%0Ax%2By%3Da%2C++xy%3Db%5C%5C%0Abegin%7Bcases%7D+a-b%3D1+%5C+a%5E2-3b%3D3+end%7Bcases%7D%26lt%3B%3D%26gt%3B+begin%7Bcases%7D+b%3Da-1+%5C+a%5E2-3%28a-1%29%3D3+end%7Bcases%7D+%26lt%3B%3D%26gt%3B%5C%0A+begin%7Bcases%7D+b%3Da-1+%5C+a%5E2-3a%3D0+end%7Bcases%7D+%26lt%3B%3D%26gt%3B%0Abegin%7Bcases%7D+b%3Da-1+%5C+a%28a-3%29%3D0+end%7Bcases%7D+%5C%0Aa_1%3D0++%3D%26gt%3Bb_1%3D0-1%3D-1%5C%0Aa_2%3D3++%3D%26gt%3Bb_2%3D3-1%3D2%5C "begin{cases} (x+y)-xy=1 \ (x+y)^2-2xy-xy=3 end{cases} <=>
begin{cases} (x+y)-xy=1 \ (x+y)^2-3xy=3 end{cases}\\
x+y=a, xy=b\\
begin{cases} a-b=1 \ a^2-3b=3 end{cases}<=> begin{cases} b=a-1 \ a^2-3(a-1)=3 end{cases} <=>\
begin{cases} b=a-1 \ a^2-3a=0 end{cases} <=>
begin{cases} b=a-1 \ a(a-3)=0 end{cases} \
a_1=0 =>b_1=0-1=-1\
a_2=3 =>b_2=3-1=2\")
Возвратимся к х и у:
1)
2)
![begin{cases} x+y=3 \ xy=2 end{cases}<=>begin{cases} x=3-y \ (3-y)y=2 end{cases}<=>begin{cases} x=3-y \ y^2-3y+2=0 end{cases}\
begin{cases} y_3=1 \ x_3=2 end{cases} begin{cases} y_4=2 \ x_4=1 end{cases}](https://tex.z-dn.net/?f=begin%7Bcases%7D+x%2By%3D3+%5C+xy%3D2+end%7Bcases%7D%26lt%3B%3D%26gt%3Bbegin%7Bcases%7D+x%3D3-y+%5C+%283-y%29y%3D2+end%7Bcases%7D%26lt%3B%3D%26gt%3Bbegin%7Bcases%7D+x%3D3-y+%5C+y%5E2-3y%2B2%3D0+end%7Bcases%7D%5C%0Abegin%7Bcases%7D+y_3%3D1+%5C+x_3%3D2+end%7Bcases%7D+begin%7Bcases%7D+y_4%3D2+%5C+x_4%3D1+end%7Bcases%7D "begin{cases} x+y=3 \ xy=2 end{cases}<=>begin{cases} x=3-y \ (3-y)y=2 end{cases}<=>begin{cases} x=3-y \ y^2-3y+2=0 end{cases}\
begin{cases} y_3=1 \ x_3=2 end{cases} begin{cases} y_4=2 \ x_4=1 end{cases}")
Ответ: (1; -1), (-1; 1), (1; 2), (2; 1).
Подставим в систему:
Возвратимся к х и у:
1)
2)
Ответ: (1; -1), (-1; 1), (1; 2), (2; 1).
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