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Zaejka
@Zaejka
June 2022
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разложите на множители:
B^6-4B^4+12B^2-9
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Answers & Comments
LyubaAlexandorovna
Пусть в^2=x, тогда x^3-4*x^2+12*x-9=(общее кратное для последних постоянных 3, делитель=х-3)
x^3-4*x^2+12*x-9|x-3 (делим столбиком, как обычно)
-x^3-3*x^2 x^2-x+3
-x^2+12*x
- -x^2+3*x
9*x-9
-9*x-9
0
x^3-4*x^2+12*x-9=(x-3)*(x^2-x+3)=(подставляем в)=(в^2-3)*(в^4-в^2+3)
2 votes
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Answers & Comments
x^3-4*x^2+12*x-9|x-3 (делим столбиком, как обычно)
-x^3-3*x^2 x^2-x+3
-x^2+12*x
- -x^2+3*x
9*x-9
-9*x-9
0
x^3-4*x^2+12*x-9=(x-3)*(x^2-x+3)=(подставляем в)=(в^2-3)*(в^4-в^2+3)