Question

125^x-25^x+(4*25^x-20)/5^x-5<=4

Answer 1

5^x=a
a³-a²+(4a²-20)/(a-5) -4≤0
(a^4-5a³-a³+5a²+4a²-20-4a+20)/(a-5)≤0
(a^4-6a³+10a²-4a)/(a-5)≤0
a(a³-6a²+10a-4)/(a-5)≤0
a[a²(a-2)-4a(a-2)+2(a-2)]/(a-5)≤0
a(a-2)(a²-4a+2)/(a-5)≤0
a=0
a-2=0⇒a=2
a²-4a+2=0
D=16-8=8
a1=(4-2√2)/2=2-√2
a2=2+√2
a-5=0⇒a=5
           _                +              _                +              _              +
--------------[0]-----------[2-√2]---------[2]---------[2+√2]--------(5)-----------
a<0⇒5^x<0нет решения
2-√2≤a≤2⇒2-√2≤5^x≤2⇒log(5)(2-√2)≤x≤log(5)2
2+√2≤a<5⇒2+√2≤5^x<5⇒log(5)(2+√2)≤x<1
x∈[log(5)(2-√2);log(5)2] U [log(5)(2+√2);1)