Đáp án + Giải thích các bước giải:

Bài $\textbf{14}:$

$\textbf{a}\bigg)$

$\textbf{A} = \dfrac{2 + x}{2 - x} - \dfrac{4x^2}{x^2 - 4} - \dfrac{2 - x}{2 + x}$

ĐKXĐ: $\begin {cases} 2 - x \ne 0 \\ x^2 - 4 \ne 0 \\ 2 + x \ne 0 \end {cases}$

$\begin {cases} x \ne 2 \\ x \ne \pm 2 \\ x \ne -2 \end {cases}$

Kết hợp các điều kiện lại, ta được $x \ne \pm 2$

Ta có:

$\textbf{A} = \dfrac{2 + x}{2 - x} - \dfrac{4x^2}{(x + 2)(x - 2)} - \dfrac{2 - x}{2 + x}$

$\phantom{\textbf{A}} = \dfrac{x + 2}{2 - x} + \dfrac{4x^2}{(x + 2)(2 - x)} + \dfrac{x - 2}{x + 2}$

$\phantom{\textbf{A}} = \dfrac{(x + 2)^2}{(x + 2)(2 - x)} + \dfrac{4x^2}{(x + 2)(2 - x)} + \dfrac{(x - 2)(2 - x)}{(x + 2)(2 - x)}$

$\phantom{\textbf{A}} = \dfrac{(x + 2)^2 + 4x^2 + (x - 2)(2 - x)}{(x + 2)(2 - x)}$

$\phantom{\textbf{A}} = \dfrac{x^2 + 4x + 4 + 4x^2 - (x - 2)^2}{(x + 2)(2 - x)}$

$\phantom{\textbf{A}} = \dfrac{5x^2 + 4x + 4 - x^2 + 4x - 4}{(x + 2)(2 - x)}$

$\phantom{\textbf{A}} = \dfrac{4x^2 + 8x}{(x + 2)(2 - x)}$

$\phantom{\textbf{A}} = \dfrac{4x(x + 2)}{(x + 2)(2 - x)}$

$\phantom{\textbf{A}} = \dfrac{4x}{2 - x}$

$\textbf{b}\bigg)$

Thay $\textbf{A} = -3$, ta có:

$\dfrac{4x}{2 - x} = -3$

$4x = -3(2 - x)$

$4x = 3x - 6$

$x = -6 (\text{tm})$

Vậy $x = -6$ thì $\textbf{A} = -3$

Bài $\textbf{15}:$

$\textbf{a}\bigg)$

$\textbf{A} = \dfrac{3 + x}{3 - x} - \dfrac{4x^2}{x^2 - 9} - \dfrac{3 - x}{3 + x}$

ĐKXĐ: $\begin {cases} 3 - x \ne 0 \\ x^2 - 9 \ne 0 \\ 3 + x \ne 0 \end {cases}$

$\begin {cases} x \ne 3 \\ x \ne \pm 3 \\ x \ne -3 \end {cases}$

Kết hợp các điều kiện lại, ta được $x \ne \pm 3$

Ta có:

$\textbf{A} = \dfrac{3 + x}{3 - x} - \dfrac{4x^2}{(x + 3)(x - 3)} - \dfrac{3 - x}{3 + x}$

$\phantom{\textbf{A}} = \dfrac{x + 3}{3 - x} + \dfrac{4x^2}{(x + 3)(3 - x)} + \dfrac{x - 3}{x + 3}$

$\phantom{\textbf{A}} = \dfrac{(x + 3)^2}{(x + 3)(3 - x)} + \dfrac{4x^2}{(x + 3)(3 - x)} + \dfrac{(x - 3)(3 - x)}{(x + 3)(3 - x)}$

$\phantom{\textbf{A}} = \dfrac{(x + 3)^2 + 4x^2 + (x - 3)(3 - x)}{(x + 3)(3 - x)}$

$\phantom{\textbf{A}} = \dfrac{x^2 + 6x + 9 + 4x^2 - (x - 3)^2}{(x + 3)(3 - x)}$

$\phantom{\textbf{A}} = \dfrac{5x^2 + 6x + 9 - x^2 + 6x - 9}{(x + 3)(3 - x)}$

$\phantom{\textbf{A}} = \dfrac{4x^2 + 12x}{(x + 3)(3 - x)}$

$\phantom{\textbf{A}} = \dfrac{4x(x + 3)}{(x + 3)(3 - x)}$

$\phantom{\textbf{A}} = \dfrac{4x}{3 - x}$

$\textbf{b}\bigg)$

Thay $\textbf{A} = -3$, ta có:

$\dfrac{4x}{3 - x} = -3$

$4x = -3(3 - x)$

$4x = 3x - 9$

$x = -9 (\text{tm})$

Vậy $x = -9$ thì $\textbf{A} = -3$

Bài $\textbf{18}:$

$\textbf{a}\bigg)$

$\textbf{A} = \dfrac{x - 2}{x + 2} - \dfrac{x}{x - 2} - \dfrac{9x + 2}{4 - x^2}$

ĐKXĐ: $\begin {cases} x + 2 \ne 0 \\ x - 2 \ne 0 \\ 4 - x^2 \ne 0 \end {cases}$

$\begin {cases} x \ne -2 \\ x \ne 2 \\ x \ne \pm 2  \end {cases}$

Kết hợp các điều kiện lại, ta được $x \ne \pm 2$

Ta có:

$\textbf{A} = \dfrac{x - 2}{x + 2} - \dfrac{x}{x - 2} + \dfrac{9x + 2}{x^2 - 4}$

$\phantom{\textbf{A}} =  \dfrac{x - 2}{x + 2} - \dfrac{x}{x - 2} + \dfrac{9x + 2}{x^2 - 4}$

$\phantom{\textbf{A}} =  \dfrac{x - 2}{x + 2} - \dfrac{x}{x - 2} + \dfrac{9x + 2}{(x + 2)(x -2 )}$

$\phantom{\textbf{A}} =  \dfrac{(x - 2)^2}{(x + 2)(x - 2)} - \dfrac{x(x + 2)}{(x + 2)(x - 2)} + \dfrac{9x + 2}{(x + 2)(x -2 )}$

$\phantom{\textbf{A}} =  \dfrac{(x - 2)^2 - x(x + 2) + 9x + 2}{(x + 2)(x - 2)}$

$\phantom{\textbf{A}} =  \dfrac{x^2 - 4x + 4 - x^2 - 2x + 9x + 2}{(x + 2)(x - 2)}$

$\phantom{\textbf{A}} =  \dfrac{3x + 6}{(x + 2)(x - 2)}$

$\phantom{\textbf{A}} =  \dfrac{3}{x - 2}$

$\textbf{b}\bigg)$

Thay $x = 3$ vào $\textbf{M}$, ta có:

$\textbf{A} = \dfrac{3}{3 - 2} = 3$

Vậy $\textbf{A} = 3$ khi $x = 3$

$\textbf{c}\bigg)$

Để $\textbf{A} = \dfrac{3}{x - 2}$ nguyên thì $3$ chia hết cho $x - 2$

Suy ra $x - 2 \in \{-3; -1; 1; 3\}$

$x \in \{-1; 1; 3; 5\}$

Vậy với $x \in \{-1; 1; 3; 5\}$ thì $\textbf{A}$ nguyên

Bài $\textbf{22}:$

$\textbf{a}\bigg)$

$\textbf{M} = \dfrac{x + 3}{x - 2} + \dfrac{x - 3}{x + 2} - \dfrac{2x^2 + 3x + 6}{x^2 - 4}$

ĐKXĐ: $\begin {cases} x - 2 \ne 0 \\ x + 2 \ne 0 \\ 4 - x^2 \ne 0 \end {cases}$

$\begin {cases} x  \ne 2 \\ x \ne -2 \\ x \ne \pm 2 \end {cases}$

Kết hợp các điều kiện lại, ta được $x \ne \pm 2$

$\textbf{M} = \dfrac{x + 3}{x - 2} + \dfrac{x - 3}{x + 2} - \dfrac{2x^2 + 3x + 6}{x^2 - 4}$

$\phantom{\textbf{M}} = \dfrac{x + 3}{x - 2} + \dfrac{x - 3}{x + 2} - \dfrac{2x^2 + 3x + 6}{(x + 2)(x - 2)}$

$\phantom{\textbf{M}} = \dfrac{(x + 3)(x + 2)}{(x + 2)(x - 2)} - \dfrac{(x - 3)(x - 2)}{(x + 2)(x - 2)} - \dfrac{2x^2 + 3x + 6}{(x + 2)(x - 2)}$

$\phantom{\textbf{M}} = \dfrac{(x + 3)(x + 2) + (x - 3)(x - 2) - 2x^2 - 3x - 6}{(x + 2)(x - 2)}$

$\phantom{\textbf{M}} = \dfrac{x^2 + 5x + 6 + x^2 - 5x + 6 - 2x^2 - 3x - 6}{(x + 2)(x - 2)}$

$\phantom{\textbf{M}} = \dfrac{-3x + 6}{(x + 2)(x - 2)}$

$\phantom{\textbf{M}} = \dfrac{-3(x - 2)}{(x + 2)(x - 2)}$

$\phantom{\textbf{M}} = \dfrac{-3}{x +2}$

$\textbf{b}\bigg)$

Thay $x = 3$ vào $\textbf{M}$, ta có:

$\textbf{M} = \dfrac{-3}{3 + 2} = -\dfrac{3}{5}$

Vậy $\textbf{M} =-\dfrac{3}{5}$ khi $x = 3$