Question

Решите уравнение 1)x(x-4)+x-4=0
2)t(t+7)-4t-28=0 ​

Answer 1

Ответ:

x^2-4x+x-4=0

x^2-3x-4=0

x¹+x²=3

x¹×x²= -4 => x¹= -1 x²= 4

t^2+7t-4t-28=0

t^2+3t-28=0

D=9-4×(-28)=9+112=121 корень из 121=11

x¹= -3-11/2 = - 14/2= -7

x²= -3+11/2= 8/2=4

Answer 2

$1)\\x(x-4)+x-4=0\\(x-4)(x+1)=0\\x-4=0\\x_1=4\\\\x+1=0\\x_2=-1\\\\x_1=4 \ \ \ x_2=-1\\\\2)\\t(t+7)-4t-28=0\\t(t+7)-4(t+7)=0\\(t+7)(t-4)=0\\t+7=0\\t_1=-7\\\\t-4=0\\t_2=4\\\\t_1=-7 \ \ \ t_2=4$