2x^2<3
2x^2=3
x=±sqrt(3/2)
x∈(-sqrt(3/2));(sqrt(3/2))
3x³-3y³+5x²-5y²=
=3(x-y)(x²+xy+y²)+5(x-y)(x+y)=
=(x-y)(3(x²+xy+y²)+5(x+y))
F(x)=3x²/2-5x+c
F(4)=3*8-20+c=4+c
4+c=10 → c=6
Ответ
F(x)=<span>3x²/2-5x+6</span>
Ответ:
1)8х-(7х+8)=9
8х-7х-8=9
8х-7х=9+8
х=17
2)5(х-3)-2(х-7)+7(2х+6)=7
5х-15-2х+14+14х+42=7
17х+41=7
17х=7-41
17х=-34
х=-2