Explicație pas cu pas:
a)
[tex]3x - 9 = 0 < = > 3x = 9 \\ = > x = 3 \\ x \in \mathbb{N};x \not \in \mathbb{Z};x \not \in \mathbb{Q};x \not \in \mathbb{R}[/tex]
b)
[tex]2x + 4 = 0 < = > 2x = - 4 \\ = > x = - 2 \\ x \not \in \mathbb{N};x \in \mathbb{Z};x \in \mathbb{Q};x \in \mathbb{R}[/tex]
c)
[tex] \frac{2}{5}x - \frac{1}{3} = 0 < = > \frac{6x - 5}{15} = 0 \\ 6x - 5 = 0 < = > 6x = 5 \\ = > x = \frac{5}{6} \\ x \not \in \mathbb{N};x \not \in \mathbb{Z};x \in \mathbb{Q};x \in \mathbb{R}[/tex]
d)
[tex] \sqrt{3}x + 2 = 0 < = > \sqrt{3}x = - 2 \\ x = - \frac{2}{ \sqrt{3} } < = > x = - \frac{2 \sqrt{3} }{3} \\ x \not \in \mathbb{N};x \not \in \mathbb{Z};x \not \in \mathbb{Q};x \in \mathbb{R}[/tex]
e)
[tex]3,12x - 1 = 0 <=> 3,12x = 1 \\ x = \frac{1}{3,12} < = > x = \frac{25}{78} \\ x \not \in \mathbb{N};x \not \in \mathbb{Z};x \in \mathbb{Q};x \in \mathbb{R}[/tex]
f)
[tex]0,(3)x + \frac{2}{5} = 0 < = > \frac{3x}{9} + \frac{2}{5} = 0 \\ \frac{x}{3} + \frac{2}{5} = 0 < = > \frac{5x + 6}{15} = 0 \\ 5x + 6 = 0 < = > 5x = - 6 \\ = > x = - \frac{6}{5} \\ x \not \in \mathbb{N};x \not \in \mathbb{Z};x \in \mathbb{Q};x \in \mathbb{R}[/tex]
g)
[tex]\sqrt{2}x - 1= \sqrt{2} - x < = > \sqrt{2}x + x = \sqrt{2} + 1 \\ x( \sqrt{2} + 1) = \sqrt{2} + 1 < = > x = \frac{\sqrt{2} + 1}{\sqrt{2} + 1} \\ = > x = 1 \\ x \in \mathbb{N};x \in \mathbb{Z};x \in \mathbb{Q};x \in \mathbb{R}[/tex]
