Question

Đã trả lời bởi chuyên gia

Answer 1

a) Rút gọn P. Với x ≠ 3, x ≠ −3, x ≠ −7.

$P = (\frac{x^{2} + 1}{x^{2} - 9} - \frac{x}{x + 3} + \frac{5}{3 - x}) : (\frac{2 x + 10}{x + 3} - 1)$

$= [\frac{x^{2} + 1}{(x + 3) (x - 3)} - \frac{x}{x + 3} - \frac{5}{x - 3}] : (\frac{2 x + 10}{x + 3} - \frac{x + 3}{x + 3})$

$= [\frac{x^{2} + 1}{(x + 3) (x - 3)} - \frac{x (x - 3)}{(x + 3) (x - 3)} - \frac{5 (x + 3)}{(x + 3) (x - 3)}] : (\frac{2 x + 10}{x + 3} - \frac{x + 3}{x + 3})$

$= \frac{x^{2} + 1 - x (x - 3) - 5 (x + 3)}{(x + 3) (x - 3)} : \frac{2 x + 10 - (x + 3)}{x + 3}$

$= \frac{x^{2} + 1 - x^{2} + 3 x - 5 x - 15}{(x + 3) (x - 3)} : \frac{x + 7}{x + 3}$

$= \frac{- 2 x - 14}{(x + 3) (x - 3)} . \frac{x + 3}{x + 7}$ $= \frac{- 2}{x - 3}$

b) Ta có |x – 1| = 2.

$\Leftrightarrow$ x – 1 = 2 hoặc x – 1 = – 2

$\Leftrightarrow$ x = 3 (loại) hoặc x = – 1 (TM).

Thay x = – 1 vào $P = \frac{- 2}{x - 3}$ , ta được:

$P = \frac{- 2}{- 1 - 3} = \frac{- 2}{- 4} = \frac{1}{2}$ .

c) Ta có $P = \frac{x + 5}{6} \Leftrightarrow \frac{- 2}{x - 3} = \frac{x + 5}{6}$

$\Leftrightarrow$ (x – 3)(x + 5) = −12

$\Leftrightarrow$ x² + 2x – 15 = −12

$\Leftrightarrow$ x² + 2x – 3 = 0

$\Leftrightarrow$ x² – x + 3x – 3 = 0

$\Leftrightarrow$ x(x – 1) + 3(x – 1) = 0

$\Leftrightarrow$ (x – 1)(x + 3) = 0

$\Leftrightarrow$ x – 1 = 0 hoặc x + 3 = 0

$\Leftrightarrow$ x = 1 (TM) hoặc x = – 3 (loại).

Vậy để $P = \frac{x + 5}{6}$ thì x = – 1.

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Answer 2