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x+y =1
x⁴ +y⁴ =17
Симметричные уравнения
* * * Известно : (x+y)⁴ =x⁴ +4x³y +6x²y² +4xy²+y⁴ * * *
{x+y =1; (x+y)⁴ -4x³y -4xy³ -6x²y² =17.
{x+y =1;(x+y)⁴ -4xy(x²+y²) -6x²y² =17 .
{x+y =1;(x+y)⁴ -4xy ((x+y)² -2xy ) -6(xy)² =17 .
{x + y =1 ; 1 -4xy(1- 2xy) -6(xy)² =17 .
1 -4xy(1- 2xy) -6(xy)² =17 .
1 -4xy+8(xy)² -6(xy)² =17 .
2(xy)² - 4xy -16 =0 .
(xy)² - 2xy -8 =0 .
(xy)₁ = - 2;
(xy)₂ = 4 ;
a) { x+y =1; xy = -2 ⇔t² -t -2 =0 * * * x² -x -2 =0 или y² -y -2 =0 * * *
t₁ = -1 ;t₂ =2.
x₁ = -1 ; y₁ =2 или x₂ =2 ; y₂ = -1 .
(-1; 2) или (2 ;-1)
б) { x+y =1; xy =4=0 ⇔t² -t +4 =0 не имеет решения .
ответ : (-1; 2) , (2 ;-1)
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