ОДЗ
{(x+3)/(x-3)>0⇒x<-3 U x>3
{x-3>0⇒x>3
{x+3>0⇒x>-3
x∈(3;∞)
перейдем к основанию 2
log(2)4/log(2)[(x+3)/(x-3)]=2(log(2)(x-3)/log(2)(1/2)-log(2)√(x+3)/log(2)(1/√2))
2/log(2)[(x+3)/(x-3)]=2(-log(2)(x-3)-1/2log(2)(x+3)/(-1/2))
2/log(2)[(x+3)/(x-3)]=2(log(2)[(x+3)/(x-3)]
log(2)[(x+3)/(x-3)]=t
2/t=2t
2t²=2
t²=1
t1=-1 U t2=1
log(2)[(x+3)/(x-3)]=-1
(x+3)/(x-3)=1/2⇒2x+6=x-3⇒x=-9∉ОДЗ
log(2)[(x+3)/(x-3)]=1
(x+3)/(x-3)=2⇒x+3=2x-6⇒x=9
Ответ х=9
Sin α=-(1-(15/17)^2α)^(1/2)=-(1-(225/289))^(1/2)=-(64/289)^(1/2)=-(8/17)
<span>tg α=-(-18/17*17/15)=-(-8/15)=8/15</span>
5x >= 2x + 9 5x - 2x >= 9 3x >= 9 x >= 3
10 - 2x <= 8 -2x <= 8 - 10 - 2x <=- 2 x >= 1
Ответ: [3; + бесконечности)
2) - 5 < 1 - (2-x)/3< 5
- 15 < 3 - 2 + x < 15
- 15 < 1 + x < 15
-16 < x < 14
Ответ: (- 16; 14)
12х*2-4х-12х*2-16х=5
-20х=5
х=-4