ЕкарныйБабай ОДЗ
x>0,x≠1
(2+1/log(5)x)*log²(5)x≤1
2log²(5)x+log(5)x≤1
log(5)x=a
2a²+a-1≤0
D=1+8=9
a1=(-1-3)/4=-1
a2=(-1+3)/4=1/2
-1≤a≤1/2
-1≤log(5)x≤1/2
1/5≤x≤√5
x∈[1/5;1) U (1;√5]

sedinalana

Verified answer

X>0 x≠1
(2 + logx 5 )log²5 x ≤1
(2+ 1/log5 x )*log²5 x ≤1
2log²5x + log5x -1 ≤0
log5x=t
2t²+t-1≤0
D=1+8=9=3²
t12=(-1+-3)/4=-1 +1/2
(t+1)(2t-1)≤0
++++++ -1 ------------- 1/2 ++++++
t⊂[-1 1/2]
log5 x ≥-1
x≥1/5
log5 x≤1/2
x≤√5
учтем одз
x⊂[1/5 1 ) U ( 1 √5]