Пошаговое объяснение:
1)х3=t
3t2+7t–6=0
D=49–4·3·(–6)=49+72=121
t=–3; t=2/3
x3 < –3 или х3 > 2/3
x < ∛–3 или х > ∛2/3
2)(2·(x–1)·x+3·(x–1)–2·x)/(x·(x–1)) > 0;
(2x2–x–3)/x·(x–1) > 0;
2x2–x–3=0
D=1+24=25
x=–1; x=3/2
__+__ (–1) __–__ (0) __+__ (1) _–_ (3/2) _+__
О т в е т. (–∞;–1)U(0;1)U(3/2;+∞)
3) x2+3x+3 > 0 при любом х, так как D=9–4·3 < 0
x2+3x+24 > 4x2+12x+12
3x2+9x–12 < 0
x2+3x–4 < 0
D=9+16=25
x=–4; x=1
О т в е т. (–4;1)
4)x2–8x–9 < 3x2+5x+2;
2x2+13x+11 > 0
D=169–88=81
x=–5,5; x=–1
О т в е т. (–∞;–5,5)U(–1;+ ∞)
