6 4X 1 2 3 X 3X 14 + + − = + (1) + ≥ ≥ −4X 1 0 X 1 1 ⇔ ⇔ − ≤ ≤4 X 3 − ≥ ĐK
Bài 4. PT : 6 4x 1 2 3 x 3x 14 + + − = + (1)
+ ≥ ≥ −4x 1 0 x 1 1 ⇔ ⇔ − ≤ ≤4 x 3 − ≥ ĐK:
(*)
3 x 0 x 3 4 ≤(1) ⇔ 3x 14 6 4x 1 2 3 x 0 + − + − − =
⇔
(4x + 1) – 2. 3. 4x 1 + + 9 + (3 – x) – 2
3 x−+ 1 = 0
⇔
(
4x 1 3+ −) (
2
+ 3 x 1− −)
2
=0 + − =⇔