Răspuns:
Bună,
[tex]a) \: \frac{x + 1}{3} - \frac{1}{2} \geqslant \frac{x}{6} \\ \frac{2x + 2}{6} - \frac{3}{6} \geqslant \frac{x}{6} | \times 6 \\ 2x + 2 - 3 \geqslant x \\ 2x - 1 \geqslant x \\ x \geqslant 1 \\ \\ [/tex]
x€ [1, + ∞)
[tex]x \sqrt{2} - 1 < x \sqrt{8} + | - 1| \\ x \sqrt{2} - 1 < 2x \sqrt{2} + 1 \\ x \sqrt{2} - 2x \sqrt{2} < 1 + 1 \\ - x \sqrt{2} < 2 \\ x > - \frac{2}{ \sqrt{2} } \\ x > - \frac{2 \sqrt{2} }{2} \\ x > - \sqrt{2} [/tex]
x€ (-2 , +∞)
[tex] \frac{2x}{ - 3} + \frac{1}{ - 4} < - \frac{x}{ - 6} + \frac{5}{ - 12} \\ \frac{8x}{ - 12} + \frac{3}{ - 12} < - \frac{2x}{ - 12} + \frac{5}{ - 12} \\ 8x + 3 > - 2x + 5 \\ 10x > 2 \\ x > \frac{2}{10} \\ x > \frac{1}{5} [/tex]
x€( 1/5 , +∞)
[tex]3 \times (x - \frac{1}{3} ) - 2 \times (x + \frac{1}{2}) \leqslant 4 \times (x - \frac{1}{4} ) \\ 3x - 1 - 2x - 1 \leqslant 4x - 1 \\ x - 4x \leqslant - 1 + 2 \\ \\ - 3x \leqslant 1 \\ x \geqslant - \frac{1}{3} [/tex]
x€ [-1/3 , +∞)
Sper că te-am ajutat.❤️
