Question

va rog multtttttt multtttttt ​

Answer 1

Explicație pas cu pas:

${5}^{ | {x}^{2} - 3x + 2| } < {25}^{x + 1}$

${5}^{ | {x}^{2} - 3x + 2| } < {( {5}^{2} )}^{x + 1}$

${5}^{ |{x}^{2} - 3x + 2| } < {5}^{2(x + 1)}$

$| {x}^{2} - 3x + 2| < 2(x + 1)$

${x}^{2} - 3x + 2 = 0 \\ (x - 1)(x - 2) = 0$

dacă:

$x \in \Big(1 ; 2 \Big)$

$- ({x}^{2} - 3x + 2) < 2(x + 1) \\ - {x}^{2} + 3x - 2 <2x + 2 \\ {x}^{2} - x + 4 > 0 , \ \Delta < 0 \implies x \in \mathbb{R}$

$\implies x \in \Big(1 ; 2 \Big) \cap \mathbb{R} = \Big(1 ; 2 \Big)$

dacă:

$x \in \Big(-\infty ;1 \Big] \cup \Big[2 ; +\infty \Big)$

${x}^{2} - 3x + 2 < 2(x + 1) \\ {x}^{2} - 3x + 2 - 2x - 2 < 0 \\ {x}^{2} - 5x < 0 \iff x(x - 5) < 0 \\ \implies x \in \Big(0 ;5 \Big)$

$\implies x \in \Big(0 ;5 \Big) \cap \ \Big( \Big(-\infty ;1 \Big] \cup \Big[2 ; +\infty \Big) \Big) = \Big(0 ;1 \Big] \cup \Big[2 ;5 \Big)$

unim cele două soluții obținute:

$\implies x \in \Big(1 ; 2 \Big) \cup \Big(0 ;1 \Big] \cup \Big[2 ;5 \Big) = \Big(0 ; 5 \Big)$