1) y'= 3^x * ln3 -2
2)y'=5^6x-1 * ln5*(6x-1)' +e^3x * (3x)' = 5^6x-1 * 6ln5 + 3e^3x
<span>3)y=log3*(x^2+2x+4)
y'= 1/[(x</span>²+2x+4)*ln3] *(2x+2)
<span>
4)y=ln*(x^2-3x)+cos3x
y'= 1/(x</span><span>²-3x) * (2x-3) -sin3x * 3</span>
<span>(16x-27y-20)²=0
(5x+18y-41,5)²</span>=0
<span>16x-27y-20=0
5x+18y-41,5=0</span>
<span>16x-27y=20 *2 32x-54y=40
5x+18y=41,5</span> *3 15x+54y=124,5
47x=164,5 x=3,5 y=4/3
Х-2/х^2-1=1-2х/х^2-1
х-2=1-2х
х-2+2х=1
3х-2=1
3х=1+2
3х=3
х=3:3
х=1
7+2x-2=>3+4x
2x-4x=>3-7+2
-2x=>-2
x<=1
Ответ:(от бесконечности; 1]