Вопрос пользователя:
1) 5t(t-5)+2=2+5t
2)1-6(x2-x-1)=7
3)u-2(u2+u-1)=2
4)5×2-2x(x-3)=0
Илюха отвечает:
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)5×2-2x(x-3)=0
5x^2 – 2x^2 +6x = 0
3x^2 +6x = 0
3x (x + 2) =
x1 = 0, x2 = -2