` c) ` Ta có:
` (x^2 + 2x - 3)/(x^2 - 1) ` `(ĐK: x ne ±1)`
` = \frac{(x^2 + 2x + 1) - 2^2}{(x - 1)(x + 1)} `
` = \frac{(x + 1)^2 - 2^2}{(x - 1)(x + 1)} `
` = \frac{(x - 1)(x + 3)}{(x - 1)(x + 1)} `
` = \frac{x + 3}{x + 1} `
Để ` C = 4 ` thì:
` \frac{x + 3}{x + 1} = 4 `
` <=> x + 3 = 4x + 4 `
` <=> x - 4x = 4 - 3 `
` <=> -3x = 1 `
` <=> x = -1/3 `
Vậy ` x = -1/3 ` thì ` C = 4 `
` d) ` Ta có:
` C = (x + 3)/(x + 1) = \frac{x + 1 + 2}{x + 1} = 1 + \frac{2}{x + 1} `
Để ` C in Z ` thì:
` 2 \vdots (x + 1) ` `(ĐK: x ne ±1)`
` => (x + 1) in Ư(2) = {±1 ; ±2} `
Ta có bảng giá trị:
$\begin{array}{|c|c|c|}\hline x+1&-2&-1&1&2\\\hline x&-3&-2&0&1\\\hline &tm&tm&tm&ktm\\\hline \end{array}$
Vậy ` x in {-3 ; -2 ; 0} ` thì ` C in Z `