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

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6 августа, 19:10

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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Тая · Answer 1

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

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

5t^2 - 30t = 0

5t (t - 6) = 0

t1 = 0, t2 = 6

2) 1-6 (x2-x-1) = 7

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

- 6x^2 + 6x = 0

-6x (x - 1) = 0

x1 = 0, x2 = 1

3) u-2 (u2+u-1) = 2

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

- 2u^2 - u = 0

-u (2u + 1) = 0

u1 = 0, u2 = - 0.5

4) 5x2-2x (x-3) = 0

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

3x^2 + 6x = 0

3x (x + 2) =

x1 = 0, x2 = - 2

1) 5t (t-5) + 2=2+5t2) 1-6 (x2-x-1) = 73) u-2 (u2+u-1) = 24) 5x2-2x (x-3) = 0

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