<span>5^2x-5*5^x*2^x-6*2^2x≤0/2^2x
(5/2)^2x-5*(5/2)^x-6≤0
(5/2)^x=a
a²-5a-6≤0
a1+a2=5Ua1*a2=-6⇒a1=-1 U a2=6
-1≤a≤6⇒-1≤(2,5)^x≤6⇒x≤log(2,5)6
x∈(-∞;log(2,5)6]
2
2^x=a
(7-2a)/(a²-12a+32)≥1/4
(28-8a-a²+12a-32)/(a²-12a+32)≥0
(a²-4a+4)/(a²-12a+32)≤0
(a-2)²/(a-4)(a-8)≤0
a=2 a=4 a=8
+ + _ +
------------[2]---------(4)------------(8)----------------------
4<a<8 U a=2
4<2^x<8 U 2^x=2
2<x<3 U x=1
x∈(2;3) U {1}</span>
Пусть комплексное число имеет вид: z = x + iy
Модуль комплексного числа: ![|z|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}](https://tex.z-dn.net/?f=%7Cz%7C%3D%5Csqrt%7B%28-3%29%5E2%2B%28-2%29%5E2%7D%3D%5Csqrt%7B13%7D)
Так как x , y < 0, то угол α ∈ III четверти, тогда
![\alpha=\rm \pi+arctg\bigg|\dfrac{y}{x}\bigg|=\pi+arctg\dfrac{2}{3}](https://tex.z-dn.net/?f=%5Calpha%3D%5Crm+%5Cpi%2Barctg%5Cbigg%7C%5Cdfrac%7By%7D%7Bx%7D%5Cbigg%7C%3D%5Cpi%2Barctg%5Cdfrac%7B2%7D%7B3%7D)
![\rm z=-3-2i=\sqrt{13}\left[\cos\bigg(arctg\dfrac{2}{3}+\pi\bigg)+i\sin\bigg(arctg\dfrac{2}{3}+\pi\bigg)\right]](https://tex.z-dn.net/?f=%5Crm+z%3D-3-2i%3D%5Csqrt%7B13%7D%5Cleft%5B%5Ccos%5Cbigg%28arctg%5Cdfrac%7B2%7D%7B3%7D%2B%5Cpi%5Cbigg%29%2Bi%5Csin%5Cbigg%28arctg%5Cdfrac%7B2%7D%7B3%7D%2B%5Cpi%5Cbigg%29%5Cright%5D)