Відповіді на вправи: 522 (1)

522. Розв’яжіть систему рівнянь:

1) $\frac{1}{x}$ – $\frac{3}{y}$ = 9;

$\frac{2}{x}$ + $\frac{1}{y}$ = 4;

/$\frac{1}{x}$ = a, $\frac{1}{y}$ = b/

a – 3b = 9| • (–2);

2a + b = 4;

+ –2a + 6b = –18; 7b = –14; b = –2; b = –2.

2a + b = 4; b = –2; a – 3 • (–2) = 9; a = 3.

$\frac{1}{x}$ = 3; x = $\frac{1}{3}$;

$\frac{1}{y}$ = –2; y = –$\frac{1}{2}$.

Відповідь: ($\frac{1}{3}$; –$\frac{1}{2}$).

2) $\frac{1}{х}$ + $\frac{1}{y}$ = $\frac{5}{6}$;

$\frac{1}{x^2}$ + $\frac{1}{y^2}$ = $\frac{13}{36}$;

/$\frac{1}{x}$ = a; $\frac{1}{y}$ = b/

a + b = $\frac{5}{6}$; a + b = $\frac{5}{6}$;

a² + b² = $\frac{13}{36}$; (a + b)² – 2ab = $\frac{13}{36}$;

($\frac{5}{6}$)² – 2ab = $\frac{13}{36}$;

2ab = $\frac{25}{36}$ – $\frac{13}{36}$;

2ab = $\frac{12}{36}$;

2ab = $\frac{1}{3}$;

ab = $\frac{1}{6}$;

a + b = $\frac{5}{6}$;

ab = $\frac{1}{6}$;

a = $\frac{1}{3}$; a = $\frac{1}{2}$;

b = $\frac{1}{2}$; b = $\frac{1}{3}$;

$\frac{1}{x}$ = $\frac{1}{3}$; $\frac{1}{x}$ = $\frac{1}{2}$;

$\frac{1}{y}$ = $\frac{1}{2}$; $\frac{1}{y}$ = $\frac{1}{3}$;

x = 3; x = 2;

y = 2; y = 3.

Відповідь: (3; 2), (2; 3).

3) x + $\frac{1}{x+y}$ = –1;

$\frac{x}{x+y}$ = –2;

/x = a; x + y = b/

a + $\frac{1}{6}$ = –1;

$\frac{a}{b}$ = –2;

a = –2b;

–2b + $\frac{1}{b}$ = –1| • b (b ≠ 0);

–2b² + 1 + b = 0| • (–1);

–2b² – b – 1 = 0;

D = (–1)² – 4 • 2 • (–1) = 9 = 3²;

b₁ = $\frac{1-3}{4}$ = $\frac{-2}{4}$ = –$\frac{1}{2}$; b₂ = $\frac{1+3}{4}$ = $\frac{4}{4}$ = 1.

b = –$\frac{1}{2}$; b = 1;

a = –2 • (–$\frac{1}{2}$); a = –2 • 1;

b = –$\frac{1}{2}$; b = 1;

a = 1; a = –2;

x = 1; x = –2;

y = –$\frac{1}{2}$ – 1; y = 1 + 2;

x = 1; x = –2;

y = –1,5; y = 3.

Відповідь: (1; –1,5), (–2; 3).

4) xy – $\frac{y}{x}$ = $\frac{1}{2}$ | • (–1);

xy – $\frac{x}{y}$ = 2;

–xy + $\frac{y}{x}$ = –$\frac{1}{2}$;

xy – $\frac{x}{y}$ = 2;

$\frac{y}{x}$ – $\frac{x}{y}$ = $\frac{3}{2}$;

/$\frac{x}{y}$ = t/ ОДЗ: x ≠ 0, y ≠ 0;

$\frac{1}{t}$ – t = $\frac{3}{2}$ |• 2t;

2 – 2t² – 3t = 0;

–2t² – 3t + 2 = 0;

D = (–3)² – 4 • (–2) • 2 = 9 + 16 = 25 = 5²;

t₁ = $\frac{3-5}{-4}$ = $\frac{-2}{-4}$ = $\frac{1}{2}$; t₂ = $\frac{3+5}{-4}$ = $\frac{8}{4}$ = –2.

$\frac{x}{y}$ = $\frac{1}{2}$; y = 2x; x • 2x – 2 = $\frac{1}{2}$; 2x2 = $\frac{5}{2}$; x2 = $\frac{5}{4}$; x1 = –$\frac{\sqrt{5}}{2}$, x2 = $\frac{\sqrt{5}}{2}$;

$\frac{x}{y}$ = –2; y = –$\frac{x}{2}$; x • (–$\frac{x}{2}$) + $\frac{1}{2}$ = $\frac{1}{2}$; –$\frac{x^2}{2}$ = 0; x = 0 – не задовольняє ОДЗ.

х = –$\frac{\sqrt{5}}{2}$; x = $\frac{\sqrt{5}}{2}$;

y = –$\sqrt{5}$; y = $\sqrt{5}$;

Відповідь: (–$\frac{\sqrt{5}}{2}$; $\sqrt{5}$), ($\frac{\sqrt{5}}{2}$; $\sqrt{5}$).