Đáp án:
Bạn tham khảo nhé
Giải thích các bước giải:
a) ĐKXĐ : `x \ne 3 \text{ và } x \ne -3`
b) `P = \frac{3}{x + 3} + \frac{1}{x - 3} - \frac{18}{9 - x^2}`
`P = \frac{3}{x + 3} + \frac{1}{x - 3} - \frac{-18}{x^2 - 9} ` $\\$ ` = \frac{3}{x + 3} + \frac{1}{x - 3} - \frac{-18}{(x + 3)(x - 3)}` $\\$ `MTC : (x + 3)(x - 3)` $\\$ `= \frac{3(x - 3)}{(x + 3)(x - 3)} + \frac{1(x + 3)}{(x + 3)(x - 3)} - \frac{-18}{(x + 3)(x - 3)}` $\\$ `= \frac{3(x - 3) + 1(x + 3) + 18}{(x + 3)(x - 3)}` $\\$ `= \frac{3x - 9 + x + 3 + 18}{(x + 3)(x - 3)} = \frac{4x + 12}{(x + 3)(x - 3)} = \frac{4(x + 3)}{(x + 3)(x - 3)} = \frac{4}{x - 3}`
c) Với P = 4 thì `P = \frac{4}{x - 3} => \frac{4}{x - 3} = 4 => 4(x - 3) = 4 => x - 3 = 1 => x = 4`
d) `\frac{4}{x - 3} => x - 3 \in Ư(4) = {1;-1;2;-2;4;-4} => x \in {4;2;5;1;7;-1}`
