Предмет: Алгебра, автор: Akhtanov02Madik

ПОМОГИТЕ, СРОЧНО НУЖНО, ДАЮ 45 БАЛЛОВ!!!
#1. Найдите наименьшие целое число, являющееся решением неравенства: (х-8)^3+24х^2>=х^2+64х.

#2. Докажите тождество:
(Х^3-9у^4)^2-(х^3+9у^4)^2+36х^3(у4-х)=-36.

Ответы

Автор ответа: Удачник66
16

1) Раскрываем скобки

x^3 - 3*8x^2 + 3*8^2x - 8^3 + 24x^2 >= x^2 + 64x

x^3 + 192x - 512 >= x^2 + 64x

x^3 - x^2 + 128x - 512 >= 0

Обозначим левую часть f(x).

f(3) = 27 - 9 + 384 - 512 = 18 - 128 = - 110 < 0

f(4) = 64 - 16 + 512 - 512 = 48 > 0

Наименьшее целое, удовлетворяющее неравенству, равно 4.

2) Вы не дописали, это выражение равно - 36x^4

(x^3 - 9y^4)^2 - (x^3 + 9y^4)^2 + 36x^3*(y^4 - x) =

= (x^3-9y^4-x^3-9y^4) (x^3-9y^4+x^3+9y^4) + 36x^3*y^4 - 36x^4 =

= - 18y^4*2x^3 + 2*18x^3*y^4 - 36x^4 = - 36x^4

Доказано.

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