Відповіді на вправи: 493 (1-4)

1) у = $\sqrt{{-x}^2+8+2x}$;

D(y): –x² + 8 + 2x $\ge$ 0;

x² – 2x – 8 $\le$ 0;

x² – 2x – 8 = 0;

D = (–2)² – 4 • (–8) = 36 = 6²;

x₁ = $\frac{2-6}{2}$ = $\frac{-4}{2}$ = –2;

x₂ = $\frac{2+6}{2}$ = $\frac{8}{2}$ = 4;

x $\in$ [–2; 4].

2) у = $\sqrt{x^2-4x-12}$;

D(y): x² – 4x – 12 $\ge$ 0;

x² – 4x – 12 = 0;

x₁ = –2, x₂ = 6;

х $\in$ (–$\infty$; –2] $\cup$ [6; +$\infty$).

3) y = $\sqrt{(x+2)(x+3)-2}$;

D(y): x + 2x + 3 – 2 $\ge$ 0;

x² + 3х + 2x + 6 – 2 $\ge$ 0;

x² + 3х + 2x + 6 – 2 = 0;

x² + 5x + 4 = 0;

x₁ = 1, x₂ = 4;

х $\in$ (–$\infty$; 1] $\cup$ [4; +$\infty$).

4) y = $\sqrt{x^2+7x-8}$;

D(y): x² + 7x – 8 $\ge$ 0;

x² + 7x – 8 = 0;

x₁ = 8, x₂ = 1;

х $\in$ (–$\infty$; –8] $\cup$ [1; +$\infty$).

5) y = $\sqrt{4x^2-13x+9}$;

D(y): 4x² – 13x + 9 $\ge$ 0;

4x² – 13x + 9 = 0;

D = (–13)² – 4 • 4 • 9 = 25 = 5²;

x₁ = $\frac{13-5}{8}$ = $\frac{8}{8}$ = 1;

x₂ = $\frac{13+5}{8}$ = $\frac{18}{8}$ = $\frac{9}{4}$ = 2,25;

х $\in$ (–$\infty$; 1] $\cup$ [2,25; +$\infty$).

6) y = $\sqrt{3(x^2+1)-10x}$;

D(y): 3(x² + 1) – 10x $\ge$ 0;

3x² + 3 – 10x $\ge$ 0;

3x² – 10x + 3 = 0;

D = (–10)² – 4 • 3 • 3 = 64 = 8²;

х₁ = $\frac{10-8}{6}$ = $\frac{2}{6}$ = $\frac{1}{3}$;

x₂ = $\frac{10+8}{6}$ = $\frac{18}{6}$ = 3.

х $\in$ (–$\infty$;$\frac{1}{3}$] $\cup$ [3; +$\infty$).