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Жанетта2012
@Жанетта2012
September 2021
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Помогите очень вас прошу на завтра
Решить неравенство
[tex]log^2_{0,2}(5x^2)+ \frac{2log^2_{0,2}x-9log_{0,2}x+2}{log_{0,2}(25x)} \leq 0 [/tex]
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sedinalana
Verified answer
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
+ + _ +
----------[0]----------------(2)----------------[2,5]--------------------
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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Verified answer
ОДЗ x>0log²(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
+ + _ +
----------[0]----------------(2)----------------[2,5]--------------------
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}