Question

Sa se rezolve in Q:

g) $\frac{1}{2}$|x+1|=5

j) $\frac{|3x-1|}{2}$-$\frac{1}{3}$=0

k) $\frac{|x+1|}{5}$-$\frac{2|x+1|}{3}$=-1$\frac{2}{3}$

Va rog, rapid, dau coroana

Answer 1

Explicație pas cu pas:

g)

$\frac{1}{2}|x+1|=5 \\ |x+1| = 10$

$x + 1 = 10 = > x = 9 \\ x + 1 = - 10 = > x = - 11$

j)

$\frac{|3x-1|}{2} - \frac{1}{3}=0 \\ \frac{|3x-1|}{2} = \frac{1}{3} \\ |3x - 1| = \frac{2}{3}$

$3x - 1 = \frac{2}{3} \\ 3x = \frac{2}{3} + 1 \\ 3x = \frac{5}{3} = > x = \frac{5}{9}$

$3x - 1 = - \frac{2}{3} \\ 3x = - \frac{2}{3} + 1 \\ 3x = \frac{1}{3} = > x = \frac{1}{9}$

k)

$\frac{|x+1|}{5}-\frac{2|x+1|}{3}=-1\frac{2}{3} \\ \frac{|x+1|}{5}-\frac{2|x+1|}{3}=-\frac{5}{3} \\ 3 |x + 1| - 10 |x + 1| = - 25 \\ - 7 |x + 1| = - 25 \\ |x + 1| = \frac{25}{7}$

$x + 1 = \frac{25}{7} \\ x = \frac{25}{7} - 1 = > x = \frac{18}{7}$

$x + 1 = - \frac{25}{7} \\ x = - \frac{25}{7} - 1 = > x = - \frac{32}{7}$