Question

Rezolvati ecuatiile cu modul: a) |2x+3|=5 b)|x-1|=|x+2|​

Answer 1

a)

$|2x+3|=5$

$2x+3= - 5 \iff 2x = - 5 - 3$

$2x = - 8 \ \ \big |:2 \iff x = - 4$

$2x+3= 5 \iff 2x = 5 - 3$

$2x = 2 \ \ \big |:2 \iff x = 1$

$\implies S = \Big\{-4;1\Big\}$

b)

$|x-1|=|x+2|$

explicităm modulele:

$x - 1 = 0 \iff x = 1$

$|x - 1| = \left\big \{\begin{matrix} - (x - 1),x < 1 \\x - 1 \ \ \ \ \ \ \ , x \geqslant 1\end{matrix}\right.$

și

$|x + 2| = 0 \iff x = - 2$

$|x + 2| = \left\big \{\begin{matrix} - (x + 2),x < - 2 \\x + 2 \ \ \ \ \ \ \ , x \geqslant - 2\end{matrix}\right.$

avem trei intervale:

x<-2; -2≤x<1; x≥1

▪︎pentru x < -2

$- (x-1)= - (x+2) \iff - x + 1 = - x - 2$

$x - x = - 2 - 1 \iff 0 \cdot x = - 3 \to f \breve {a}r \breve {a}\ \ solu \c {t}ie$

▪︎pentru -2 ≤ x < 1

$- (x - 1) = x + 2 \iff - x + 1 = x + 2$

$x + x = 2 - 1 \iff 2x = 1 \ \ \big|:2$

$\implies x = \dfrac{1}{2}$

▪︎pentru x ≥ 1

$x - 1= x + 2 \iff x - x = 2 + 1$

$0 \cdot x = - 3 \to f \breve {a}r \breve {a}\ \ solu \c {t}ie$

$\implies S = \Big\{ \dfrac{1}{2} \Big\}$