Question

Поясните как решать это....

Answer 1

2
ОДЗ x>0
(2+4log(4)x+11)/(log²(4)x-9)≥-1
(4log(4)x+13)/(log²(4)x-9)+1≥0
(4log(4)x+13+log²(4)x-9)/(log²(4)x-9)≥0
log(40x=t
(t²+4t+4)/(t²-9)≥0
(t+2)²/(t-3)(t+3)≥0
t=-2  t=3  t=-3
  +           _                    _             +
------(-3)--------[-2]----------(3)-------------
t<-3⇒log(4)x<-3⇒x<1/64
t=-2⇒log(4)x=-2⇒x=1/16
t>3⇒log(4)x>3⇒x>64
x∈(0;1/64) U (64;∞) U {1/16}
3
[2^x*(5^x-25)-2(5^x-25)]/([(x-4)(x-1)]≤0
(2^x-2)(5^x-25)/(x-4)(x-1)≤0
x=1  x=2  x=4  
  _             +                _              +
-------(1)----------[2]----------(4)----------
x∈(-∞;1) U [2;5)
4
ОДЗ
(x-3)(x+3)>0
x<-3 U x>3
log(2)(x²-9)=t
t²-9t+20≥0
t1+t2=9 U t1*t2=20
t1=4 U t2=5
t≤4⇒log(2)(x²-9)≤4⇒x²-9≤16⇒x²-25≤0⇒(x-5)(x+5)≤0⇒-5≤x≤5
t≥5⇒log(2)(x²-9)≥5⇒x²-9≥32⇒x²-41≥0⇒(x-√41)(x+√41)≥0⇒x≤-√41 U x≥√41
\\\\\\                                                                           ///////////////
--------[-√41]-----------[-5]-------[-3]--------[3]-----[5]---------[√41]--------
                                 /////////////////////////////////////////
////////////////////////////////////////////////              \\\\\\\\\\\\\\\\\\\\\\\
x∈[-5;-3] U [3;5]
5
2^x=t
t-6≤(9t-37)/(t-4)(t-3)+1/(t-4)
t-6≤(9t-37+t-3)/(t-4)(t-3)
t-6≤10(t-4)/(t-4)(t-3)
t-6≤10/(t-3),t≠4
(t-6) -10/(t-3)≤0
(t²-6t-3t+18-10)/(t-3)≤0
(t²-9t+8))/(t-3)≤0
(t-1)(t-8)/(t-3)≤0
t=1  t=8  t=3
  _               +             _                 +
------[1]-----------(3)------------[8]---------
t≤1⇒2^x≤1⇒x≤0
3<t<4⇒3<2^x<4⇒log(2)3<x<2
4,t≤8⇒4<2^x≤8⇒2<x≤3
x∈(-∞;0] U (log(2)3;2) U (2;3]