a, Điều kiện xác định: $x \neq -3, x \neq 2$
b, Ko ghi lại đề!
$=\dfrac{x+2}{x+3}-\dfrac{5}{(x-2)(x+3)}+\dfrac{1}{2-x}$
$=\dfrac{(x+2)(x-2)}{(x-2)(x+3)}-\dfrac{5}{(x-2)(x+3)}+\dfrac{1}{2-x}$
$=\dfrac{(x+2)(x-2)-5}{(x-2)(x+3)}+\dfrac{1}{2-x}$
$=\dfrac{-9+x^2}{(x-2)(x+3)}+\dfrac{1}{2-x}$
$=\dfrac{x-3}{x-2}+\dfrac{1}{2-x}$
$=\dfrac{x-3}{x-2}+\dfrac{-1}{x-2}$
$=\dfrac{x-4}{x-2}$ (1)
c, $\dfrac{x-4}{x-2}=\dfrac{-3}{4}$
$\to 4(x-4)=-3(x-2)$
$\to 4x-16=-3x+6$
$\to 7x=22$
$\to x = \dfrac{22}{7}$ (TMĐK)
d, $A=\dfrac{x-4}{x-2}=\dfrac{(x-2)-2}{x-2}=1-\dfrac{2}{x-2}$
Để biểu thức A nguyên thì `x-2 ∈ Ư(2)={1;-1;2;-2}`
`⇔x ∈ {3;1;4;0}`
e, $x^2-9=0$
$\to$ \(\left[ \begin{array}{l}x=3\\x=-3\end{array} \right.\)
Thay $x=3$ vào (1) ta có: $A=\dfrac{3-4}{3-2}=-1$
Thay $x=-3$ vào (1) ta có: $A=\dfrac{-3-4}{-3-2}=\dfrac{7}{5}$
