|7-4x|=7-4x log2(x+1)+log2x 1 - вопрос №921589074742121 от dmitriwatchdmitry21 22.07.2022 17:19

Question

Математика: |7-4x|=7-4x log2(x+1)+log2x 1

dmitriwatchdmitry21 · Answer

1) Нули модуля: 7 - 4х =04х =7х =1,75а) х 1,75; 7-4х = 7- 4х; б) х 1,75; 4х -7 = 7 - 4х; 8х = 14; х = 1 3/4ответ: х = 1,752)(х + 1)×х 2х^2 + х -2 0х^2 + х -2 =0D= 9x1= -2; X2 = 1ответ: (-2;1)...

|7-4x|=7-4x log2(x+1)+log2x < 1