1) b^1/6 (b^1/6-4) - (b^1/6-2) ^2 = 2) (b^1/8-c^1/4) (b^1/8+c^1/4) - (5b^1/8+2c^1/4) (3b^1/8-4c^1/4) =

Question

Алексей Семенов

Математика

5 апреля, 21:56

1) b^1/6 (b^1/6-4) - (b^1/6-2) ^2 = 2) (b^1/8-c^1/4) (b^1/8+c^1/4) - (5b^1/8+2c^1/4) (3b^1/8-4c^1/4) =

+2

Ответы (1)

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Ева Павлова · Answer 1

1) b^1/6 * b^1/6 - 4 * b^1/6 - (b^1/6 - 2) * (b^1/6 - 2) = b^1/6 * b^1/6 - 4 * b^1/6 - b^2/6 + 2 * b^1/6 + 2 * b^1/6 - 4 = b^ (1/6 + 1/6) - 4 * b^1/6 - b^1/3 + 4 * b^1/6 - 4 = b^1/3 - b^1/3 - 4 = - 4;

Ответ: - 4.

2) (b^1/8) ^2 - (c^1/4) ^2 - 15 * b^2/8 + 20 * c^1/4 * b^1/8 - 6 * c^1/4 * b^1/8 + 8 * c^2/4 = b^1/4 - c^1/2 - 15 * b^1/4 + 14 * c^1/4 * b^1/8 + 8 * c^1/2 = - 14 * b^1/4 + 14 * c^1/4 * b^1/8 + 7 * c^1/2;

Ответ: - 14 * b^1/4 + 14 * c^1/4 * b^1/8 + 7 * c^1/2.