1) 5t (t-5) + 2=2+5t 2) 1-6 (x2-x-1) = 7 3) u-2 (u2+u-1) = 2 4) 5x2-2x (x-3) = 0

Question

Медина Митрофанова

Математика

13 февраля, 14:17

1) 5t (t-5) + 2=2+5t 2) 1-6 (x2-x-1) = 7 3) u-2 (u2+u-1) = 2 4) 5x2-2x (x-3) = 0

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Эмилия Синицына · Answer 1

1) 5t (t - 5) + 2 = 2 + 5t.

5t^2 - 25 + 2 = 2 + 5t.

5t^2 - 5t - 25 + 2 - 2 = 0.

5t^2 - 5t - 25 = 0.

t^2 - t - 5 = 0.

Посчитаем дискриминант:

D = 1 + 20 = 21.

t1 = (1 - √21) / 2; t2 = (1 + √21) / 2.

2) 1 - 6 (x^2 - x - 1) = 7.

1 - 6x^2 + 6x + 6 - 7 = 0.

-6x^2 + 6x = 0.

x (x - 1) = 0.

x1 = 0; x2 = 1.

3) u - 2 (u^2 + u - 1) = 2.

u - 2u^2 - 2u + 2 - 2 = 0.

-2u^2 - u = 0.

u (2u + 1) = 0.

u1 = 0; u2 = - 0,5.

4) 5x^2 - 2x (x - 3) = 0.

5x^2 - 2x^2 + 6x = 0.

3x^2 + 6x = 0.

x (x + 2) = 0.

x1 = 0; x2 = - 2.