Question

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Answer 1

$+) N ế u 8 - 24 x \geq 0 < = > x \leq \frac{1}{3} t a c ó : {(3 x - 1)}^{4} = 8 - 24 x = > {(3 x - 1)}^{4} + 24 x - 8 = 0 = > {(3 x - 1)}^{4} + 8 . (3 x - 1) = 0 = > (3 x - 1) [{(3 x - 1)}^{3} + 8] = 0 = > 3 x - 1 = 0 h o ặ c {(3 x - 1)}^{3} + 8 = 0 = > 3 x = 1 h o ặ c {(3 x - 1)}^{3} = - 8 = > x = \frac{1}{3} (t m) h o ặ c 3 x - 1 = - 2 = > x = \frac{1}{3} h ặ c 3 x = - 1 = > x = \frac{1}{3} h ặ c x = \frac{- 1}{3} (t m)$

$+) N ế u 8 - 24 x < 0 = > x > \frac{1}{3} t a c ó : {(3 x - 1)}^{4} = 24 x - 8 = > {(3 x - 1)}^{4} - 8 . (3 x - 1) = 0 = > (3 x - 1) [{(3 x - 1)}^{3} - 8] = 0 = > 3 x - 1 = 0 h ặ c {(3 x - 1)}^{3} = 8 = > x = \frac{1}{3} (l o ạ i) h o ặ c 3 x - 1 = 2 = > 3 x = 3 = > x = 1 (t m) V ậ y S = \{\frac{- 1}{3}; \frac{1}{3}; 1\}$ .

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Answer 2

$+) N ế u 8 - 24 x \geq 0 < = > x \leq \frac{1}{3} t a c ó : {(3 x - 1)}^{4} = 8 - 24 x = > {(3 x - 1)}^{4} + 24 x - 8 = 0 = > {(3 x - 1)}^{4} + 8 . (3 x - 1) = 0 = > (3 x - 1) [{(3 x - 1)}^{3} + 8] = 0 = > 3 x - 1 = 0 h o ặ c {(3 x - 1)}^{3} + 8 = 0 = > 3 x = 1 h o ặ c {(3 x - 1)}^{3} = - 8 = > x = \frac{1}{3} (t m) h o ặ c 3 x - 1 = - 2 = > x = \frac{1}{3} h ặ c 3 x = - 1 = > x = \frac{1}{3} h ặ c x = \frac{- 1}{3} (t m)$

$+) N ế u 8 - 24 x < 0 = > x > \frac{1}{3} t a c ó : {(3 x - 1)}^{4} = 24 x - 8 = > {(3 x - 1)}^{4} - 8 . (3 x - 1) = 0 = > (3 x - 1) [{(3 x - 1)}^{3} - 8] = 0 = > 3 x - 1 = 0 h ặ c {(3 x - 1)}^{3} = 8 = > x = \frac{1}{3} (l o ạ i) h o ặ c 3 x - 1 = 2 = > 3 x = 3 = > x = 1 (t m) V ậ y S = \{\frac{- 1}{3}; \frac{1}{3}; 1\}$ .

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