Question

|x-3|+|x-4|=1, sa se rezolve în mulțimea numerelor reale. ​

Answer 1

Hei! :)

$|x-3|+|x-4|=1\\Rezolvam\ fiecare\ modul\ in\ parte\\x-3=0 => x=3\\x-4=0=> x=4 \\\\\\x\ \ \ \ \ \|-\ infinit\ \ \ \ 3\ \ \ \ \ \ 4\ \ \ \ \ \ \ \ +\ infinit\\\\x-3| ------0++++++++++\\x-4|---------0+++++++++\\\\=> cazul\ 1 : x\ apartine (-\ infinit;3)\\=> -x+3-x+4=1\\-2x=-6 => x=3\\\\=> cazul\ 2: x\ apartine (3;4]\\=> x-3-x+4=1 => 1=1 (nu\ are\ solutie)\\\\=> cazul\ 3: x\ apartine(4; +\ infinit]\\=> x-3+x-4=1\\2x=8 => x=4\\\\=> din\ toate\ cazurile\ x\ apartine [3,4]$

Answer 2

Răspuns:

x = [3, 4]

Explicație pas cu pas:

x - 3 = 0 ⇒ x = 3

x - 4 = 0 ⇒ x = 4

_____________

x < 3

Ix - 3I + Ix - 4I = 3 - x + x - 4 = -1 ≠ 1

x > 4

Ix - 3I + Ix - 4I = x - 3 + x - 4 = 2x - 7 > 1

x ∈ [3,  4]

Ix - 3I + Ix - 4I = x - 3 + 4 - x = 1