Повторюємо алгебру за 8 клас: 82 (1-2)

82. Скоротіть дроби:

1) $\frac{x^2-x-6}{2x+4}$ = $\frac{(x-3)(x+2)}{2(x+2)}$ = $\frac{x-3}{2}$;

x² – x – 6 = (x – 3)(x + 2);

x₁ + x₂ = 1; x₁ • x₂ = –6;

x₁ = 3; x₂ = –2.

2) $\frac{{3x}^2-8x-3}{x^2-9}$ = $\frac{(3x+1)(x-3)}{(x-3)(x+3)}$ = $\frac{3x+1}{x+3}$;

3x² – 8x – 3 = 3(x + $\frac{1}{3}$)(x – 3) = (3x + 1)(x – 3);

D = (–8)² – 4 • 3 • (–3) = 64 + 36 = 100;

x₁ = $\frac{8-\sqrt{100}}{2•3}$ = $\frac{8-10}{6}$ = $\frac{-2}{6}$ = –$\frac{1}{3}$.

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

3) $\frac{{2x}^2-3x-2}{x+1-{2x}^2}$ = $\frac{(2x+1)(x-2)}{-(2x+1)(x-1)}$ = $\frac{x-2}{1-x}$.

2x² – 3x – 2 = 2(x + $\frac{1}{2}$)(x – 2) = (2x + 1)(x – 2);

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

x₁ = $\frac{3-\sqrt{25}}{2•2}$ = $\frac{3-5}{4}$ = $\frac{-2}{4}$ = –$\frac{1}{2}$;

x₂ = $\frac{3+5}{4}$ = $\frac{8}{4}$ = 2.

–2x² + x + 1 = –2(x + $\frac{1}{2}$)(x – 1) = –(2x + 1)(x – 1).

D = 1² – 4 • (–2) • 1 = 1 + 8 = 9;

x₁ = $\frac{-1-\sqrt{9}}{2•(-2)}$ = $\frac{-1-3}{-4}$ = $\frac{-4}{-4}$ = 1;

x₂ = $\frac{-1+3}{-4}$ = $\frac{2}{-4}$ = –$\frac{1}{2}$.