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$