Penyelesaian dari |x + 3 | ≤ |x - 1| adalah....

Penyelesaian dari |x + 3 | ≤ |x - 1| adalah....

Pembahasan :

Cara 1 |x + 3| ≤ |x - 1|| |x + 3|² ≤ |x - 1|² x² + 6x + 9 ≤ x² - 2x + 1 8x ≤ -8 x ≤ -1 HP = {x | x ≤ -1, x ∈ R} Cara 2 : |a| ≤ |b| => (a + b)(a - b) ≤ 0 |x + 3| ≤ |x - 1| ((x + 3) + (x - 1)) ((x + 3) - (x - 1)) ≤ 0 (2x + 2)(4) ≤ 0 8x + 8 ≤ 0 8x ≤ -8 x ≤ -1 HP = {x | x ≤ -1, x ∈ R}