3. a)
3ˣ⁺²-2*3ˣ⁺¹+3ˣ<12<br>3²*3ˣ-2*3*3ˣ+3ˣ<12<br>9*3ˣ-6*3ˣ+3ˣ<12<br>4*3ˣ<12<br>3ˣ<3¹<br>x<1.<br>3. b)
(log₀,₅x)²+3*log₀,₅x-4≤0 ОДЗ: x>0
log₀,₅x=t
t²+3t-4≤0
t²+3t-4=0 D=25
t₁=1 log₀,₅x=1 x₁=0,5
t₂=-4 log₀.₅x=-4 x₂=16
(t+4)(t-1)≤0
t∈[-4;1]
log₀,₅x≥-4 log₀,₅x≥log₀,₅16 x≤16
log₀,₅x≤1 log₀,₅x≤log0,5 x≥0,5 ⇒
x∈[0,5;16].