Відповіді до завдань § 12 - § 14: 12.31 (1-2)

1) 2х² – 3х $\le$ 2(х – 1);

2х² – Зх – 2(х – 1) $\le$ 0;

2х² – Зх – 2х + 2 $\le$ 0;

2х² – 5х + 2 $\le$ 0;

2х² – 5х + 2 = 0;

D = 25 – 16 = 9;

x₁ = $\frac{5+3}{4}$ = 2; x₂ = $\frac{5-3}{4}$ = $\frac{1}{2}$.

$\frac{1}{2}$ $\le$ х $\le$ 2.

2) 4х(х + 2) > 5;

4x² + 8х – 5 > 0;

y = 4х² + 8х – 5;

4х² + 8х – 5 = 0;

D = 64 + 80 = 144;

x₁ = $\frac{-8-12}{8}$ = –$\frac{20}{8}$ = –2$\frac{1}{2}$;

x₂ = $\frac{-8+12}{8}$ = $\frac{4}{8}$ = $\frac{1}{2}$.

х = $\in$ (–$\infty$; –2,5) $\cup$ (0,5; +$\infty$).

3) х(х – 8) > (2х – 1)²;

x² – 8х > 4х² – 4х + 1;

Зх² + 4х + 1 < 0;

y = Зх² + 4х + 1;

Зх² + 4х + 1 = 0;

D = 16 – 12 = 4;

х₁ = $\frac{-4+2}{6}$ = –$\frac{1}{3}$;

x₂ = $\frac{-4-2}{6}$ = –1. –1 < х < –$\frac{1}{3}$.

4) Зх(х – 2) + 1 > (х – 1)²;

Зх² – 6х + 1 – х² + 2х –1 > 0;

2х² – 4х > 0;

y = 2х² – 4х; 2х² – 4х = 0;

2х(х – 2) = 0; х₁ = 0, х₂ = 2.

x $\in$ (–$\infty$; 0) $\cup$ (2; +$\infty$).