Question

Đã trả lời bởi chuyên gia

Answer 1

$a, A = \frac{x + 1}{x - 2}, đ k : x \neq 2 = \frac{x - 2 + 3}{x - 2} = 1 + \frac{3}{x - 2} A \in Z k h i x - 2 \in Ư (3) = \{- 3; - 1; 1; 3\} = > x \in \{- 1; 1; 3; 5\} (t m) v ậ y x = \{- 1; 1; 3; 5\} b, B = \frac{2 x - 1}{x + 5}, đ k : x \neq - 5 = \frac{2 x + 10 - 11}{x + 5} = 2 - \frac{11}{x + 5} B \in Z k h i x + 5 \in Ư (11) = \{- 11; - 1; 1; 11\} = > x \in \{- 16; - 6; - 4; 6\} (t m) V ậ y x = \{- 16; - 6; - 4; 6\} c, C = \frac{10 x - 9}{2 x - 3} = \frac{10 x - 15 + 6}{2 x - 3} = 5 + \frac{6}{2 x - 3} C \in Z k h i 2 x - 3 \in Ư (6) = \{- 6; - 3; - 2; - 1; 1; 2; 3; 6\} = > 2 x \in \{- 3; 0; 1; 2; 4; 5; 6; 9\} = > x \in \{- \frac{3}{2}; 0; \frac{1}{2}; 1; 2; \frac{5}{2}; 3; \frac{9}{2}\} M à x \in Z = > x = \{0; 1; 2; 3\} V ậ y x = \{0; 1; 2; 3\}$ .

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Answer 2

$a, A = \frac{x + 1}{x - 2}, đ k : x \neq 2 = \frac{x - 2 + 3}{x - 2} = 1 + \frac{3}{x - 2} A \in Z k h i x - 2 \in Ư (3) = \{- 3; - 1; 1; 3\} = > x \in \{- 1; 1; 3; 5\} (t m) v ậ y x = \{- 1; 1; 3; 5\} b, B = \frac{2 x - 1}{x + 5}, đ k : x \neq - 5 = \frac{2 x + 10 - 11}{x + 5} = 2 - \frac{11}{x + 5} B \in Z k h i x + 5 \in Ư (11) = \{- 11; - 1; 1; 11\} = > x \in \{- 16; - 6; - 4; 6\} (t m) V ậ y x = \{- 16; - 6; - 4; 6\} c, C = \frac{10 x - 9}{2 x - 3} = \frac{10 x - 15 + 6}{2 x - 3} = 5 + \frac{6}{2 x - 3} C \in Z k h i 2 x - 3 \in Ư (6) = \{- 6; - 3; - 2; - 1; 1; 2; 3; 6\} = > 2 x \in \{- 3; 0; 1; 2; 4; 5; 6; 9\} = > x \in \{- \frac{3}{2}; 0; \frac{1}{2}; 1; 2; \frac{5}{2}; 3; \frac{9}{2}\} M à x \in Z = > x = \{0; 1; 2; 3\} V ậ y x = \{0; 1; 2; 3\}$ .

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