bonjour,
(x+1)/10≥(2x-3)/5
(x+1)/10≥2(2x-3)/2x5
(x+1)/10≥4x-6/10
x+1≥4x-6
1+6≥4x-x
7≥3x
7/3≥x
(5-x)²∠100
(5-x)²-100∠0
(5-x)²-10²∠0
[(5-x)+10)(5-x)-10)]∠0
(5-x+10)(5-x-10)∠0
(15-x)(-x-5)∠0
x -∞ -5 15 +∞
15-x + + + 0 -
-x-5 + 0 - - -
(15-x)(-x-5) + 0 - 0 +
(15-x)(-x-5)∠0
si x ∈]-5; 15[
