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

Q666Princess666Q Q666Princess666Q    1   08.03.2019 23:40    2

Ответы
dmitriwatchdmitry21 dmitriwatchdmitry21  06.06.2020 22:48

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

 

 

ПОКАЗАТЬ ОТВЕТЫ
Другие вопросы по теме Алгебра