Ответы на вопрос:
Одз x> 0 log²(0,2)(5x²)=(log(0,2)5+log(0,2)x²)²=(2log(0,2)x-1)² log(0,2)(25x)=log(0,2)25+log(0,2)x=log(0,2)x-2 log(0,2)x=a (2a-1)²+(2a²-9a+2)/(a-2)≤0 [(2a-1)²(a-2)+2a²-9a+2]/(a-2)≤0 (4a³-8a²-4a²+8a+a-2+2a²-9a+2)/(a-2)≤0 (4a³-10a²)/(a-2)≤0 2a²(2a-5)/(a-2)≤0 a=0 2a-5=0⇒a=2,5 a-2=0⇒a=2 + + _ + , a=0⇒log(0,2)x=0⇒x=1 2< a≤2,5⇒2< log(0,2)x≤2,5⇒1/(25√5)≤x< 1/25 x∈[1/25√5; 1/25) u {1}
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