Question

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Answer 1

$b, {(x - 1)}^{2} - 5 = 20 = > {(x - 1)}^{2} = 20 + 5 = > {(x - 1)}^{2} = 25 = > x - 1 = 5 h o ặ c x - 1 = - 5 = > x = 1 + 5 h o ặ c x = - 5 + 1 = > x = 6 h o ặ c x = - 4 V ậ y S = \{6; - 4\} c, \frac{x}{15} = \frac{6.5}{2 x}, đ k : x \neq 0 = > 2 . x . x = 6.5.15 = > 2 x^{2} = 450 = > x^{2} = 225 = > x = \pm 15 (t / m) V ậ y x = \pm 15 d, \frac{x + 1}{2} = \frac{\frac{8}{25}}{x + 1}, đ k : x \neq - 1 = > (x + 1) (x + 1) = 2 . \frac{8}{25} = > {(x + 1)}^{2} = \frac{16}{25} = > x + 1 = \frac{4}{5} h o ặ c x + 1 = - \frac{4}{5} = > x = \frac{4}{5} - 1 h o ặ c x = - \frac{4}{5} - 1 = > x = - \frac{1}{5} h o ặ c x = - \frac{9}{5} V ậ y S = \{- \frac{1}{5}; \frac{- 9}{5}\}$

$e, |x + 1| + {(x + 2)}^{2} = 2 = > |x + 1| + x^{2} + 4 x + 4 - 2 = 0 = > |x + 1| + x^{2} + 4 x + 2 = 0 +) x + 1 \geq 0 h a y x \geq - 1 t a c ó : x + 1 + x^{2} + 4 x + 2 = 0 = > x^{2} + 5 x + 3 = 0 = > (x^{2} + 5 x + \frac{25}{4}) - \frac{13}{4} = 0 = > {(x + \frac{5}{2})}^{2} = \frac{13}{4} = > x + \frac{5}{2} = \sqrt{\frac{13}{4}} h o ặ c x + \frac{5}{2} = - \sqrt{\frac{13}{4}} = > x = \sqrt{\frac{13}{4}} - \frac{5}{2} h o ặ c x = - \sqrt{\frac{13}{4}} - \frac{5}{2} = > x = \frac{\sqrt{13} - 5}{2} (t / m) h o ặ c x = \frac{- \sqrt{13} - 5}{2} (l o ạ i) +) x + 1 < 0 h a y x < - 1 t a c ó : - x - 1 + x^{2} + 4 x + 2 = 0 = > x^{2} + 3 x + 1 = 0 = > (x^{2} + 3 x + \frac{9}{4}) - \frac{5}{4} = 0 = > {(x + \frac{3}{2})}^{2} = \frac{5}{4} = > x + \frac{3}{2} = \pm \sqrt{\frac{5}{4}} = > x = \sqrt{\frac{5}{4}} - \frac{3}{2} (l o ạ i) h o ặ c x = - \sqrt{\frac{5}{4}} - \frac{3}{2} (t / m) = > x = \frac{- \sqrt{5} - 3}{2} V ậ y S = \{\frac{- \sqrt{5} - 3}{2}; \frac{\sqrt{13} - 5}{2}\}$ .

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Answer 2

$b, {(x - 1)}^{2} - 5 = 20 = > {(x - 1)}^{2} = 20 + 5 = > {(x - 1)}^{2} = 25 = > x - 1 = 5 h o ặ c x - 1 = - 5 = > x = 1 + 5 h o ặ c x = - 5 + 1 = > x = 6 h o ặ c x = - 4 V ậ y S = \{6; - 4\} c, \frac{x}{15} = \frac{6.5}{2 x}, đ k : x \neq 0 = > 2 . x . x = 6.5.15 = > 2 x^{2} = 450 = > x^{2} = 225 = > x = \pm 15 (t / m) V ậ y x = \pm 15 d, \frac{x + 1}{2} = \frac{\frac{8}{25}}{x + 1}, đ k : x \neq - 1 = > (x + 1) (x + 1) = 2 . \frac{8}{25} = > {(x + 1)}^{2} = \frac{16}{25} = > x + 1 = \frac{4}{5} h o ặ c x + 1 = - \frac{4}{5} = > x = \frac{4}{5} - 1 h o ặ c x = - \frac{4}{5} - 1 = > x = - \frac{1}{5} h o ặ c x = - \frac{9}{5} V ậ y S = \{- \frac{1}{5}; \frac{- 9}{5}\}$

$e, |x + 1| + {(x + 2)}^{2} = 2 = > |x + 1| + x^{2} + 4 x + 4 - 2 = 0 = > |x + 1| + x^{2} + 4 x + 2 = 0 +) x + 1 \geq 0 h a y x \geq - 1 t a c ó : x + 1 + x^{2} + 4 x + 2 = 0 = > x^{2} + 5 x + 3 = 0 = > (x^{2} + 5 x + \frac{25}{4}) - \frac{13}{4} = 0 = > {(x + \frac{5}{2})}^{2} = \frac{13}{4} = > x + \frac{5}{2} = \sqrt{\frac{13}{4}} h o ặ c x + \frac{5}{2} = - \sqrt{\frac{13}{4}} = > x = \sqrt{\frac{13}{4}} - \frac{5}{2} h o ặ c x = - \sqrt{\frac{13}{4}} - \frac{5}{2} = > x = \frac{\sqrt{13} - 5}{2} (t / m) h o ặ c x = \frac{- \sqrt{13} - 5}{2} (l o ạ i) +) x + 1 < 0 h a y x < - 1 t a c ó : - x - 1 + x^{2} + 4 x + 2 = 0 = > x^{2} + 3 x + 1 = 0 = > (x^{2} + 3 x + \frac{9}{4}) - \frac{5}{4} = 0 = > {(x + \frac{3}{2})}^{2} = \frac{5}{4} = > x + \frac{3}{2} = \pm \sqrt{\frac{5}{4}} = > x = \sqrt{\frac{5}{4}} - \frac{3}{2} (l o ạ i) h o ặ c x = - \sqrt{\frac{5}{4}} - \frac{3}{2} (t / m) = > x = \frac{- \sqrt{5} - 3}{2} V ậ y S = \{\frac{- \sqrt{5} - 3}{2}; \frac{\sqrt{13} - 5}{2}\}$ .

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