Question

rezolvati ecuații în Q :3[8-5(3x-2)]+4[4x-9)=2[3(2x-5)+28]-49 ;e)2[3(x+1)-4(2x-3)]-18=11-3[2(3x-5)-3(x+4)] ;f)25-4[6(x+1)-3(3x+4)]=5[2(3x-7)-3(x-5)]+23;ajutați-ma vă rog!​

Answer 1

Răspuns:

Explicație pas cu pas:

d) 3[8 - 5(3x - 2)] + 4(4x - 9) = 2[3(2x - 5) + 28] - 49

3(8 - 15x + 10) + 16x - 36 = 2(6x - 15 + 28) - 49

3(18 - 15x) + 16x - 36 = 2(6x + 13) - 49

54 - 45x + 16x - 36 = 12x + 26 - 49

18 - 29x = 12x - 23

18 + 23 = 12x + 29x

41 = 41x

x = 41 : 41

x = 1

e)2[3(x + 1) - 4(2x - 3)] - 18 = 11 - 3[2(3x - 5) - 3(x + 4)]

2(3x + 3 - 8x + 12) - 18 = 11 - 3(6x - 10 - 3x -12)

2(15 - 5x) - 18 = 11 - 3(3x - 22)

30 - 10x - 18 = 11 - 9x + 66

12 - 10x = 77 - 9x

10x - 9x = 12 - 77

x = -65

f) 25 - 4[6(x + 1) - 3(3x + 4)] = 5[2(3x - 7) - 3(x - 5)] + 23

25 - 4(6x + 6 - 9x - 12) = 5(6x - 14 - 3x + 15) + 23

25 - 4(-3x - 6) = 5(3x + 1) + 23

25 + 12x + 24 = 15x + 5 + 23

49 + 12x = 15x + 28

49 - 28 = 15x - 12x

21 = 3x

x = 21 : 3

x = 7