Question

решите систему урайнений
x2+y2-2xy=1
X+y=3

Answer 1

x2+y2-2xy=1
x=3-y

(3-y)2+y2-2(3-y)y=1

y=1
y=2

x=2
x=1

(x1,y1)=(2,1)
(x2,y2)=(1,2)

Answer 2

$\binom{{x}^{2} + {y}^{2} - 2xy = 1}{x + y = 3}$
$\binom{(x - y {)}^{2} = 1}{x + y = 3}$
$\binom{(x - y {)}^{2} = 1}{x = 3 - y \: }$
$\binom{(3 - y - y {)}^{2} = 1}{x = 3 - y \: }$
$\binom{(3 - 2y{)}^{2} = 1}{x = 3 - y \:}$
$\binom{9 + 2 \times3 \times 2y- (2y{)}^{2} = 1}{x = 3 - y \:}$
$\binom{9 +12y- (2y{)}^{2} = 1}{x = 3 - y \:}$
$\binom{9 +12y-4{y}^{2} = 1}{x = 3 - y \:}$
$\binom{9 +12y-4{y}^{2} -1= 0}{x = 3 - y \:}$
$\binom{8+12y-4{y}^{2}= 0}{x = 3 - y \:}$
Решаем первую часть (квадратное уравнение) :
$8 +12y-4{y}^{2} =1$
$a=-4; b=12; c=.$
$D={8}^{2}-4\times(-4) \times 9 = // 64 + 36 = 100$
y1=18/2=9 ,
y2=-2/2= -1 .

x = 3 - y ,
x1 = 3-9 = -6 ,
x2 = 3+1 = 4 .
[tex]Ответ\: : \: (-6;9) , (4;-1 )