Đáp án:

a) ĐKXĐ: `x\ne(-3); x\ne2`.

b) $C= \dfrac{{x - 4}}{{x - 2}}$.

c) $x=\dfrac{{22}}{7}$.

d) `x∈{0;1;3;4}`.

e) `C=-1`.

Giải thích các bước giải:

Câu 12:

a) ĐKXĐ: \(\left\{ \begin{array}{l}
x + 3 \ne 0\\
2 - x \ne 0\\
{x^2} + x - 6 = (x + 3)(x - 2) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne  - 3\\
x \ne 2
\end{array} \right.\)

b) $A = \dfrac{{x + 2}}{{x + 3}} - \dfrac{5}{{{x^2} + x - 6}} + \dfrac{1}{{2 - x}}\\
 = \dfrac{{x + 2}}{{x + 3}} - \dfrac{5}{{(x - 2)(x + 3)}} - \dfrac{1}{{x - 2}}\\
 = \dfrac{{{x^2} - 4 - 5 - x - 3}}{{(x - 2)(x + 3)}}\\
 = \dfrac{{{x^2} - x - 12}}{{(x - 2)(x + 3)}}\\
 = \dfrac{{(x + 3)(x - 4)}}{{(x - 2)(x + 3)}}\\
 = \dfrac{{x - 4}}{{x - 2}}$

c) $A = \dfrac{{ - 3}}{4} \Leftrightarrow \dfrac{{x - 4}}{{x - 2}} = \dfrac{{ - 3}}{4} \Leftrightarrow 4x - 16 =  - 3x + 6\\ \Leftrightarrow 7x = 22 \Leftrightarrow x = \dfrac{{22}}{7}$

Vậy $x=\dfrac{{22}}{7}$ thì $A=\dfrac{{ - 3}}{4}$.

d) $A = \dfrac{{x - 4}}{{x - 2}} = 1 - \dfrac{2}{{x - 2}}$

Để `A` nguyên `=> x-2∈Ư(2)={±1;±2}`

`x-2=1=>x=3 (TM)`

`x-2=-1=>x=1 (TM)`

`x-2=2=>x=4 (TM)`

`x-2=-2=>x=0 (TM)`

Vậy `x∈{0;1;3;4}`.

e) ${x^2} - 9 = 0 \Leftrightarrow {x^2} = 9 \Leftrightarrow \left[ \begin{array}{l}
x = 3{\rm{ }}(TM)\\
x =  - 3{\rm{ }}(KTM)
\end{array} \right. \Rightarrow x = 3$

Với `x=3` $\Rightarrow C = \dfrac{{3 - 4}}{{3 - 2}} =  - 1$.