За хранение 1год можно было получить 100000 + 150\%, т.е
<span>(а-6)(а²+2)-(4а+1)(а-3)=
=a</span>³+2a-6a²-12-(4a²-12a+a-3)=
=a³+2a-6a²-12-(4a²-11a-3)=
=a³+2a-6a²-12-4a²+11a+3=
=a³+13a-10a²-9.
при а=3====>>>3<span>³+13*3-10*3²-9=
=27+39-10*9-9=
=27+39-90-9
=-33</span>
7x-2x²≥0
x(7-2x)≥0
x=0 x=3,5
0≤x≤3,5
9-4x³≠0
4x³≠9
x≠∛(9/4)
x∈[0;∛(9/4)) U (∛(9/4);3,5]