Відповіді на вправи: 520 (1)

520. Розв’яжіть систему рівнянь.

1) x² + y² + x + y = 18;

x² + y² + xy = 19;

(x + y)² – 2xy + x + y = 18;

(x + y)² – 2xy + xy = 19;

(x + y)² – 2xy + x + y = 18;

(x + y)² – xy = 19;

x + y = a; xy = b;

a² – 2b + a = 18;

a² – b = 19;

b = a² – 19;

a² – 2(a² – 19)+ a = 18;

a² – 2a² + 38 + a – 18 = 0;

–a² + a + 20 = 0;

a² – a – 20 = 0;

a = 5; a = –4;

b = 6; b = –3;

x + y = 5; x + y = –4;

xy = 6; xy = –3;

x = 2; x = 3; t² + 4t – 3 = 0;

y = 3; y = 2; D = 4² – 4 • 1 • (–3) = 16 + 12 = 28;

t₁ = $\frac{-4-\sqrt{28}}{2}$ = $\frac{-4-2\sqrt{7}}{2}$ = –2 – $\sqrt{7}$;

t₂ = $\frac{-4+\sqrt{28}}{2}$ = $\frac{-4+2\sqrt{7}}{2}$ = –2 + $\sqrt{7}$;

x = –2 – $\sqrt{7}$; x = –2 + $\sqrt{7}$;

y = –2 + $\sqrt{7}$; y = –2 – $\sqrt{7}$.

Відповідь: (2; 3), (3; 2), (–2; –$\sqrt{7}$; –2 + $\sqrt{7}$), (–2 +$\sqrt{7}$; –2 –$\sqrt{7}$).

2) x² + y² + x + y = 18; +

x² – y² + x – y = 6;

2x² + 2x = 24 |: 2;

x² + x – 12 = 0;

x₁ = –4, x₂ = 3;

x² + y² + x + y = 18;–

x² – y² + x – y = 6;

2y² + 2y = 12| : 2;

y² + y – 6 = 0;

y₁ = –3, y₂ = 2;

Відповідь: (–4; 3), (3; 2), (–4; 2), (3; –3).

3) x + y + xy = 5;

x² + y² + xy = 7;

x + y + xy = 5;

(x + y)² – 2xy + xy = 7;

x + y + xy = 5;

(x + y)² – xy = 7;

x + y = a; xy = b;

a + b = 5;

a² – b = 7;

a = 5 – b;

(5 – b)² – b = 7;

25 – 10b + b² – b – 7 = 0;

b² – 11b + 18 = 0;

a = 5 – b;

(5 – b)² – b = 7;

25 – 10b + b² – b – 7 = 0;

b² – 11b + 18 = 0;

b₁ = 2, b₂ = 9;

b = 2; b = 9;

a = 3; a = –4;

x + y = 3; t² + 4t + 9 = 0;

xy = 2; D = 4² – 4 • 1 • 9 < 0; t ∈ ∅.

x = 1; x = 2;

y = 2; y = 1.

Відповідь: (1; 2), (2; 1).

4) x² + y² – xy = 3 | • (4);

2x² + 2y² + xy = 12;

4x² + 4y² – 4xy = 12;

2x² + 2y² + xy = 12;

4x² + 4y² – 4xy = 2x² + 2y² + xy;

4x² + 4y² – 4xy – 2x² – 2y² – xy = 0;

2x² + 2y² – 5xy = 0 |: 2y² (y ≠ 0);

$\frac{{2x}^2}{{2y}^2}$ + $\frac{{2y}^2}{{2y}^2}$ – $\frac{5\mathrm{xy}}{{2y}^2}$ = 0;

$\frac{x^2}{y^2}$ + 1 – $\frac{5x}{2y}$ = 0;

/$\frac{x}{y}$ = t/; t² – 2,5t + 1 = 0;

D = (2,5)² – 4 • 1 • (1) = 6,25 – 4 = 2,25 = 1,5²;

t₁ = $\frac{\mathrm{2,5}-\mathrm{1,5}}{2}$ = $\frac{1}{2}$ = 0,5; t₂ = $\frac{\mathrm{2,5}+\mathrm{1,5}}{2}$ = $\frac{4}{2}$ = 2;

$\frac{x}{y}$ = 0,5 або $\frac{x}{y}$ = 2; y = 0,5x або y = 2x;

x2 + (0,5x)2 – x • (0,5x) = 3; x2 + 0,25x2 + 0,5x2 = 3; 1,75x2 = 3; x2 = $\frac{3}{\mathrm{1,75}}$; x1 = –$\sqrt{\frac{3}{\mathrm{1,75}}}$; x2 = $\sqrt{\frac{3}{\mathrm{1,75}}}$;	x2 + (2x)2 – x • 2x = 3; x2 + 4x2 – 2x2 = 3; 3x2 = 3; x2 = 1; x1 = –1, x2 = 1.
x = –$\sqrt{\frac{3}{\mathrm{1,75}}}$; y = $\sqrt{1\frac{5}{7}}$;	x = $\sqrt{\frac{3}{\mathrm{1,75}}}$; y = –$\sqrt{1\frac{5}{7}}$;	x = –1; y = –2;	x = 1; y = 2.
Відповідь: (–$\sqrt{\frac{3}{\mathrm{1,75}}}$; $\sqrt{1\frac{5}{7}}$), ($\sqrt{\frac{3}{\mathrm{1,75}}};-\sqrt{1\frac{5}{7}}$), (–1; –2), (1; 2).

5) x³ + y³ = 35;

x + y = 5;

(x + y)(x² – xy + y²) = 35;

x + y = 5;

5(x² – xy + y²) = 35;

x + y = 5;

x² – xy + y² = 7;

x + y = 5;

x = 5 – y;

(5 – y)² – (5 – y)y + y² = 7;

25 – 10y + y² – 5y + y² + y² – 7 = 0;

3y² – 15y + 18 = 0 |: 3;

y² – 5y + 6 = 0;

y₁ = 2, y₂ = 3;

y = 2; y = 3;

x = 5 – 2; x = 5 – 3;

y = 2; y = 3;

x = 3; x = 2.

Відповідь: (3; 2), (2; 3).

6) x³ + y³ = 28;

x² + y² – xy = 7;

(x + y)(x² + y² – xy) = 28;

x² + y² – xy = 7;

(x + y) • 7 = 28;

x² + y² – xy = 7;

x + y = 4;

x² + y² – xy = 7;

x = 4 – y;

(4 – y)² + y² – (4 – y)y = 7;

16 – 8y + y² + y² – 4y + y² – 7 = 0;

3y² – 12y + 9 = 0|: 3;

y² – 4y + 3 = 0;

y₁ = 1, y₂ = 3;

y = 1; y = 3;

x = 3; x = 1.

Відповідь: (3; 1), (1; 3).