1) D(f)=R
2)f'(x)=(12x-x^3)=12-3x^2
3) 12-3x^2=0
3x^2-12=0
3x^2=12
x^2=12/3
x^2=4
x=4 f'(x)>0 4>0
Вот как то так...........
27-5(2-6b)-14b+9
27-10+30b-14b+9
26+16b
-4(7z-23+4d)-2(d-z)
-28z+92-14d-2d+2z
-26z+92-16d
5k-(6k-(2k-3))
5k-(6k-2k+3)
5k-6k+2k-3
-9k-3
7y-(4y-((z-3y)-8z))
7y-(4y-z+3y-8z)
7y-4y+z-3y+8z
4y+z
А
{1/x- 1/y=1/6⇒6(y-x)=xy
{x-y = -1 ⇒y=x+1
6(x+1-x)=x²+x
x²+x-6=0
x1+x2=-1 U x1*x2=-6
x1=-3⇒y1=-3+1=-2
x2=2⇒y2=2+1=3
(-3;-2);(2;3)
b
{1/x+ 1/y=1/6⇒6(x+y)=xy
{x+y=1
⇒x=1-y
6-y+y²=0
D=1-24=-23 нет решения