Log₅(2x²-5x)≤log₅3
D(y): 2x²-5x>0
x(2x-5)>0
///////////// ////////////
_______o________o______
0 2,5
x∈(-∞;0)U(2,5;+∞)
т.к. основание логарифмы больше единицы (5>1), то:
2x²-5x≤3
2x²-5x-3≤0
D=5²-4*2*(-3)+25+24=49
x₁=(5+7)/2*2=3
x₂=(5-7)/2*2=-0,5
2(x-3)(x+0,5)≤0
\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\
/////////////////////////
______.___o____o____._____
-0,5 0 2,5 3
Ответ: x∈[-0,5;0)U(2,5;3]