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Korovink
@Korovink
July 2022
2
4
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Представьте в виде многочлена (1+2k)(1-2k+4k²)=
(а-1)(а²+а-1)=
(a+b)(2a-3b)=
(2b+c)(c-2b)=
(2a-1)(4a²+2a+1)=
(2k-1)(-1)=
(x²-a)(2+x)=
(2b+3)(3b-2)=
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sedinalana
(1+2k)(1-2k+4k²)=1+8k
³
(а-1)(а²+а-1)=a
³-1
(a+b)(2a-3b)=2a
²-ab-3b²
(2b+c)(c-2b)=c
²-4b²
(2a-1)(4a²+2a+1)=8a
³-1
(2k-1)(-1)=1-2k
(x²-a)(2+x)=x
³+2x²-ax-2a
(2b+3)(3b-2)=6b
²+5b-6
1 votes
Thanks 1
Azik2121
Ответы:1)1+2k³
2)a³-2a+1
3)2a²-ab-3b²
4)-4b²+c²
5)8a³-1
6)1-2k
7)2x²+x³-2a-ax
8)6b²+5b-6
2 votes
Thanks 1
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Answers & Comments
(а-1)(а²+а-1)=a³-1
(a+b)(2a-3b)=2a²-ab-3b²
(2b+c)(c-2b)=c²-4b²
(2a-1)(4a²+2a+1)=8a³-1
(2k-1)(-1)=1-2k
(x²-a)(2+x)=x³+2x²-ax-2a
(2b+3)(3b-2)=6b²+5b-6
2)a³-2a+1
3)2a²-ab-3b²
4)-4b²+c²
5)8a³-1
6)1-2k
7)2x²+x³-2a-ax
8)6b²+5b-6