Explicație pas cu pas:
a) x (x² - 1) - (x - 1)³ < 0
[tex]x( {x}^{2} - 1) - {(x - 1)}^{3} < 0 \\ x(x - 1)(x + 1) - (x - 1)({x}^{2} - 2x + 1) < 0 \\ (x - 1)( {x}^{2} + x - {x}^{2} + 2x - 1) < 0 \\ (x - 1)(3x - 1) < 0 < = > 3(x - 1)(x - \frac{1}{3}) < 0 \\ = > \\ \frac{1}{3} < x < 1 < = > x \in \left( \frac{1}{3} ; 1 \right)[/tex]
b) (3x - 2) la a doua - 4x(x + 1) ≤ 16 + (2x + 1) la a doua
[tex]{(3x - 2)}^{2} - 4x(x + 1) \leqslant 16 + {(2x + 1)}^{2} \\ [/tex]
[tex]9 {x}^{2} - 12x + 4 - 4 {x}^{2} - 4x \leqslant 16 + 4 {x}^{2} + 4x + 1 \\ [/tex]
[tex]{x}^{2} - 20x - 13 \leqslant 0[/tex]
[tex]{(x - 10)}^{2} \leqslant 113 \\ = > \\ - \sqrt{113} + 10 \leqslant x \leqslant \sqrt{113} + 10 \\ x \in \left[- \sqrt{113} + 10 ; \sqrt{113} + 10\right][/tex]
c) 2x² +3 supra 2 - 4 x( x-1) supra 3 ≥-1 supra 6
[tex]\frac{2 {x}^{2} + 3}{2} - \frac{4x(x - 1)}{3} \geqslant - \frac{1}{6} \\ 3(2 {x}^{2} + 3) - 8x(x - 1) \geqslant - 1 \\ 6 {x}^{2} + 9 - 8 {x}^{2} + 8x + 1 \geqslant 0 \\ - 2 {x}^{2} + 8x + 10 \geqslant 0 \\ {x}^{2} - 4x + 5 \leqslant 0 \\ (x + 1)(x - 5) \leqslant 0 \\ = > \\ - 1 \leqslant x \leqslant 5 < = > x \in \left[ - 1 ; 5\right][/tex]
