Tolong di jawab uji kompetensi 3 no. 1-6, Trimkaih sebelumnya

1. |x+7| = 2
√(x+7)² = 2
(x+7)² = 2²
(x+7)² - 2² = 0
(x+7+2)(x+7-2) = 0
(x+9)(x+5) = 0
maka x = -9 atau x = -5
hp={-9,-5)

2. |4-x| = 3
√(4-x)² = 3
(4-x)² = 3²
(4-x)² - 3² = 0
(4-x+3)(4-x-3) = 0
(-x+7)(-x+1) = 0
maka x = 1 atau x = 7
hp ={1,7}

3. |3x+2| = 5
√(3x+2)² = 5
(3x+2)² = 5²
(3x+2)² - 5² = 0
(3x+2+5)(3x+2-5) = 0
(3x + 7)(3x-3) = 0
x = -7/3 , x = 3/3

hp ={-7/3, 3/3}

4. |2x-1| = 7
√(2x-1)² = 7
(2x-1)² = 7²
(2x-1)² - 7² = 0
(2x-1+7)(2x-1-7) = 0
(2x+6)(2x-8) =0
x = -6/2 , x = 8/2

hp={-6/2, 8/2}

5. |8-5x| = 3
√(8-5x)² = 3
(8-5x)² = 3²
(8-5x)² - 3² = 0
(8-5x+3)(8-5x-3) = 0
(-5x+11)(-5x+5) = 0
x = 11/5 , x = 5/5

hp ={11/5 , 5/5}

6. |x-2| = |x+5|
√(x-2)² = √(x+5)²
(x-2)² = (x+5)²
(x-2)² - (x+5)² = 0
(x-2+x+5)(x-2-x-5) = 0
(2x + 3)(-7) = 0
x = -3/2 , x = 7

hp ={-3/2, 7}


semoga membantu maaf kalau salah, saya juga baru belajar:)