5(x+y)+4xy=32 xy (x+y)=12 решить методом замены переменных

Заменяем: x+y=z; xy=t;
5z+4t=32;
t*z=12;
выражаем t из 2 уравнения: t=12/z и подставляем в 1:
5z+4*(12/z)=32;
5z^2-32z+48=0; D=64; z1=4; z2=2,4;
t1=3; t2=5;
теперь составляем систему:
1)x+y=4;
xy=3;
y=3/x;
x^2-4x+3=0; D=4; x1=3; x2=1;
y1=1; y2=3; (3;1) и (1;3);
ответ: (3;1) и (1;3)

5x³ -  3x² - 3x + 5 = 0

5x³ +5 -  3x² - 3x  = 5(x³ + 1) - 3x(x + 1) = 5(x + 1)(x² - x + 1) -3x(x + 1) = (x + 1)(5x² -5x + 5 - 3x) = (x + 1)(5x² - 8x + 5) = 0

x = -1

5x² - 8x + 5 = 0

D = 64 - 80 < 0

x ∈ ∅ при x ∈ R

ответ -1

(x + 1/x)² - 5(x + 1/x) + 6 = 0

x ≠ 0

x + 1/x = t

t² - 5t + 6 = 0

D = 25 - 24 = 1

t12 = (5 +- 1)/2 = 2   3

1. t = 2

x + 1/x = 2

(x² - 2x + 1)/x = 0

(x - 1)²/ x = 0

x = 1

2. t = 3

x + 1/x = 3

(x² - 3x + 1)/x = 0

D = 9 - 4 = 5

x12 = (3 +- √5)/2

ответ (3 +- √5)/2, 1

x⁴ - 5x³ + 8x² - 5x + 1 = 0

x ≠ 0

разделим на x²

1/x² + x² = 1/x² + 2*x²*1/x² + x² - 2*x²*1/x² = (x + 1/x)² - 2

x² - 5x + 8 - 5/x + 1/x² = x² + 1/x² - 5(x + 1/x) + 8 = (x + 1/x)² - 2 - 5(x + 1/x) + 8 = (x + 1/x)² - 5(x + 1/x) + 6 = 0

x + 1/x = t

t² - 5t + 6 = 0

это уравнение было номер 2

D = 25 - 24 = 1

t12 = (5 +- 1)/2 = 2   3

1. t = 2

x + 1/x = 2

(x² - 2x + 1)/x = 0

(x - 1)²/ x = 0

x = 1

2. t = 3

x + 1/x = 3

(x² - 3x + 1)/x = 0

D = 9 - 4 = 5

x12 = (3 +- √5)/2

ответ (3 +- √5)/2, 1

Объяснение:

А) 5x^3 - 3x^2 - 3x + 5 = 0

5x^3 + 5 - 3x^2 - 3x = 0

5(x^3 + 1) - 3x(x + 1) = 0

5(x + 1)(x^2 - x + 1) - 3x(x + 1) = 0

(x + 1)(5x^2 - 5x + 5 - 3x) = 0

(x + 1)(5x^2 - 8x + 5) = 0

x + 1 = 0 => x = -1

5x^2 - 8x + 5 = 0

D = 8^2 - 4 * 5 * 5 = 64 - 100 = -36

∅

ответ: x = -1

Б) (x + 1/x)^2 - 5(x + 1/x) + 6 = 0

t = x + 1/x

t^2 - 5t + 6 = 0

D = 5^2 - 4 * 1 * 6 = 25 - 24 = 1

t1 = (5 + 1) / 2 = 6/2 = 3

t2 = (5 - 1) / 2 = 4/2 = 2

x + 1/x = 3

x^2 - 3x + 1 = 0

D = 3^2 - 4 * 1 * 1 = 9 - 4 = 5

x1 = (3 - √5) / 2

x2 = (3 + √5) / 2

x + 1/x = 2

x^2 - 2x + 1 = 0

D = 2^2 - 4 * 1 * 1 = 4 - 4 = 0

x = 2 / 2 = 1

ответ: x1 = (3 - √5) / 2 ; x2 = 1 ; x3 = (3 + √5) / 2.