Bantuin jawab dong tentang penyelesaian tiap pertidaksamaan nilai mutlak​

a.)

|5x-3| < |1-3x|

(5x-3 – (1-3x)).(5x-3 + (1-3x)) < 0

(5x-3-1+3x).(5x-3+1-3x) < 0

(8x-4).(2x-2) < 0

Maka,

8x-4 = 0

8x = 4

x = 4/8

x = 1/2

Atau

2x-2 = 0

2x = 2

x = 2/2

x = 1

Diminta < 0

Sehingga

1/2 < x < 1

Hp = {x | 1/2 < x < 1, x anggota bilangan Real}

b.)

|x+2| ≥ 3|x-4|

|x+2| ≥ |3x-12|

(x+2 – (3x-12)).(x+2 + (3x-12)) ≥ 0

(x+2-3x+12).(x+2+3x-12) ≥ 0

(-2x+14).(4x-10) ≥ 0

Maka,

-2x+14 = 0

-2x = -14

x = -14/-2

x = 7

Atau

4x-10 = 0

4x = 10

x = 10/4

x = 5/2

Diminta ≥ 0

Sehingga

5/2 ≤  x ≤  7

Hp = {x | 5/2 ≤  x ≤  7, x anggota bilangan Real}

c.)

|1-5x| ≤ |3x-4|

(1-5x – (3x-4)).(1-5x + (3x-4)) ≤ 0

(1-5x-3x+4).(1-5x+3x-4) ≤ 0

(-8x+5).(-2x-3) ≤ 0

Maka,

-8x+5 = 0

-8x = -5

x = -5/-8

x = 5/8

Atau

-2x-3 = 0

-2x = 3

x = 3/-2

x = -3/2

Diminta ≤ 0

Sehingga

-3/2 ≤ x ≤ 5/8

Hp = {x | -3/2 ≤ x ≤ 5/8, x anggota bilangan Real}

d.)

|3x-2| > |x+2|

(3x-2 – (x+2)).(3x-2 + (x+2)) > 0

(3x-2-x-2).(3x-2+x+2) > 0

(2x-4).(4x) > 0

Maka,

2x-4 = 0

2x = 4

x = 4/2

x = 2

Atau

4x = 0

x = 0/4

x = 0

Diminta > 0

Sehingga

x < 0 atau x > 2

Hp = {x | x < 0 atau x > 2, x anggota bilangan Real}

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