Đáp án:

$\begin{array}{l}
2)a)\left( {x + 2} \right)\left( {x + 3} \right) - \left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Rightarrow \left( {x + 2} \right)\left( {x + 3 - x + 2} \right) = 0\\
 \Rightarrow \left( {x + 2} \right).5 = 0\\
 \Rightarrow x + 2 = 0\\
 \Rightarrow x =  - 2\\
Vậy\,x =  - 2\\
b)4{x^2} - 9 - \left( {3x + 1} \right)\left( {2x - 3} \right) = 0\\
 \Rightarrow \left( {2x - 3} \right)\left( {2x + 3} \right) - \left( {3x + 1} \right)\left( {2x - 3} \right) = 0\\
 \Rightarrow \left( {2x - 3} \right)\left( {2x + 3 - 3x - 1} \right) = 0\\
 \Rightarrow \left( {2x - 3} \right)\left( {2 - x} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = 2
\end{array} \right.\\
Vậy\,x = 2;x = \dfrac{3}{2}\\
c){x^3} + 3{x^2} + 3x + 1 = 26 + 1\\
 \Rightarrow {\left( {x + 1} \right)^3} = 27\\
 \Rightarrow x + 1 = 3\\
 \Rightarrow x = 3 - 1\\
 \Rightarrow x = 2\\
Vậy\,x = 2\\
B3)\\
a)Dkxd:x \ne 2;x \ne 3\\
C = \dfrac{{2x - 9}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} - \dfrac{{x + 3}}{{x - 2}} + \dfrac{{2x + 1}}{{x - 3}}\\
 = \dfrac{{2x - 9 - \left( {x + 3} \right)\left( {x - 3} \right) + \left( {2x + 1} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{2x - 9 - \left( {{x^2} - 9} \right) + 2{x^2} - 3x - 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{2x - 9 - {x^2} + 9 + 2{x^2} - 3x - 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} - x - 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{x + 1}}{{x - 3}}\\
b)C =  - \dfrac{1}{2}\\
 \Rightarrow \dfrac{{x + 1}}{{x - 3}} = \dfrac{{ - 1}}{2}\\
 \Rightarrow 2\left( {x + 1} \right) = 3 - x\\
 \Rightarrow 2x + 2 = 3 - x\\
 \Rightarrow 2x + x = 3 - 2\\
 \Rightarrow 3x = 1\\
 \Rightarrow x = \dfrac{1}{3}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{1}{3}\\
c){x^2} - 4 = 0\\
 \Rightarrow {x^2} = 4\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x =  - 2\left( {tm} \right)
\end{array} \right.\\
Thay\,x =  - 2\\
 \Rightarrow C = \dfrac{{x + 1}}{{x - 3}} = \dfrac{{ - 2 + 1}}{{ - 2 - 3}} = \dfrac{1}{5}\\
d)C > 0\\
 \Rightarrow \dfrac{{x + 1}}{{x - 3}} > 0\\
 \Rightarrow \left[ \begin{array}{l}
x > 3\\
x <  - 1
\end{array} \right.\\
Vậy\,x > 3;x <  - 1
\end{array}$