Làm nhanh giúp mình với ạ mình đang cần gấp

Đáp án:

$\begin{array}{l}
a)a)Dkxd:\left\{ \begin{array}{l}
x + 2 \ne 0\\
x - 2 \ne 0\\
4 - {x^2} \ne 0\\
x + 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne  - 2\\
x \ne  - 3
\end{array} \right.\\
Vậy\,x \ne 2;x \ne  - 2;x \ne  - 3\\
b)\\
B = \left( {\dfrac{x}{{x + 2}} + \dfrac{2}{{x - 2}} - \dfrac{{4x}}{{4 - {x^2}}}} \right).\dfrac{{x - 2}}{{x + 3}}\\
 = \dfrac{{x\left( {x - 2} \right) + 2\left( {x + 2} \right) + 4x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x - 2}}{{x + 3}}\\
 = \dfrac{{{x^2} - 2x + 2x + 4 + 4x}}{{x + 2}}.\dfrac{1}{{x + 3}}\\
 = \dfrac{{{x^2} + 4x + 4}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{x + 2}}{{x + 3}}\\
c)2x - 6 = 0\\
 \Leftrightarrow x = 3\left( {tm} \right)\\
 \Leftrightarrow B = \dfrac{{x + 2}}{{x + 3}} = \dfrac{{3 + 2}}{{3 + 3}} = \dfrac{5}{6}\\
d)B = 4\\
 \Leftrightarrow \dfrac{{x + 2}}{{x + 3}} = 4\\
 \Leftrightarrow x + 2 = 4x + 12\\
 \Leftrightarrow 3x =  - 10\\
 \Leftrightarrow x =  - \dfrac{{10}}{3}\left( {tm} \right)\\
Vậy\,x =  - \dfrac{{10}}{3}
\end{array}$