Explicație pas cu pas:
[tex]|2x+1|+|-4x-2|+|3x+5|<1 \\ [/tex]
[tex]2x + 1 = 0 < = > 2x = - 1 = > x = - \frac{1}{2} \\ [/tex]
[tex]x < 0 = > |2x + 1| = - 2x - 1 \\ x \geqslant 0 = > |2x + 1| = 2x + 1[/tex]
[tex]-4x - 2 = 0 < = > 4x = - 2 = > x = - \frac{1}{2} \\ [/tex]
[tex]x < - \frac{1}{2} = > |-4x - 2| = - 4x - 2 \\ x \geqslant - \frac{1}{2} = > |-4x - 2| = 4x + 2[/tex]
[tex]3x + 5 = 0 < = > 3x = - 5 = > x = - \frac{5}{3} \\ [/tex]
[tex]x < - \frac{5}{3} = > |3x + 5| = - 3x - 5 \\ x \geqslant - \frac{5}{3} = > |3x + 5| = 3x + 5 [/tex]
1)
[tex]x < - \frac{5}{3} \\ [/tex]
[tex] - 2x - 1 - 4x - 2 - 3x - 5 < 1 \\ - 9x - 8 < 1 < = > - 9x < 9 \\ = > x > - 1[/tex]
=> x ∈ Ø
2)
[tex]- \frac{5}{3} < x < - \frac{1}{2} \\ [/tex]
[tex] - 2x - 1 - 4x - 2 + 3x + 5 < 1 \\ - 3x + 2 < 1 < = > - 3x < - 1 \\ = > x > \frac{1}{3}[/tex]
=> x ∈ Ø
3)
[tex]x > - \frac{1}{2} \\ [/tex]
[tex]2x + 1 + 4x + 2 + 3x + 5 < 1 \\ 9x + 8 < 1 < = > 9x < - 7 \\ = > x < - \frac{7}{9}[/tex]
=> x ∈ Ø
→ inecuația nu are soluții în mulțimea numerelor reale
