Đáp án:

a) \(\dfrac{{2x}}{{2x - 3}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne  \pm \dfrac{3}{2}\\
B = \dfrac{{3x\left( {2x - 3} \right) - 4\left( {2x + 3} \right) - 4{x^2} + 23x + 12}}{{\left( {2x + 3} \right)\left( {2x - 3} \right)}}.\dfrac{{2x + 3}}{{x + 3}}\\
 = \dfrac{{6{x^2} - 9x - 8x - 12 - 4{x^2} + 23x + 12}}{{\left( {2x + 3} \right)\left( {2x - 3} \right)}}.\dfrac{{2x + 3}}{{x + 3}}\\
 = \dfrac{{2{x^2} + 6x}}{{\left( {2x + 3} \right)\left( {2x - 3} \right)}}.\dfrac{{2x + 3}}{{x + 3}}\\
 = \dfrac{{2x\left( {x + 3} \right)}}{{\left( {2x + 3} \right)\left( {2x - 3} \right)}}.\dfrac{{2x + 3}}{{x + 3}}\\
 = \dfrac{{2x}}{{2x - 3}}\\
b)2{x^2} + 7x + 3 = 0\\
 \to \left( {2x + 1} \right)\left( {x + 3} \right) = 0\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x =  - 3
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x =  - 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
B = \dfrac{{2.\left( { - \dfrac{1}{2}} \right)}}{{2\left( { - \dfrac{1}{2}} \right) - 3}} = \dfrac{1}{4}\\
B = \dfrac{{2.\left( { - 3} \right)}}{{2\left( { - 3} \right) - 3}} = \dfrac{2}{3}
\end{array} \right.\\
c)B = \dfrac{{2x}}{{2x - 3}} = \dfrac{{2x - 3 + 3}}{{2x - 3}} = 1 + \dfrac{3}{{2x - 3}}\\
B \in Z \to \dfrac{3}{{2x - 3}} \in Z \to 2x - 3 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
2x - 3 = 3\\
2x - 3 =  - 3\\
2x - 3 = 1\\
2x - 3 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = 0\\
x = 2\\
x = 1
\end{array} \right.\\
d)\left| B \right| < 1\\
 \to  - 1 < B < 1\\
 \to \left\{ \begin{array}{l}
 - 1 < \dfrac{{2x}}{{2x - 3}}\\
\dfrac{{2x}}{{2x - 3}} < 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{2x + 2x - 3}}{{2x - 3}} > 0\\
\dfrac{{2x - 2x + 3}}{{2x - 3}} < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
2x - 3 < 0\\
4x - 3 < 0
\end{array} \right.\\
 \to x < \dfrac{3}{4}
\end{array}\)

e) Để B đạt GTNN

⇔ \(\dfrac{3}{{2x - 3}}\) đạt GTNN

⇔ \({2x - 3}\) đạt GTLN

\(2x - 3 = 9 \to x = 6 \to MinB = 1 + \dfrac{3}{{2.6 - 3}} = \dfrac{4}{3}\)

( Thay biểu thức chữ E thành B b nha )

a, B = ($\frac{3x}{2x+3}$ + $\frac{4}{3-2x}$ - $\frac{4x^{2}-23x-12}{4x^{2}-9}$) : ($\frac{x+3}{2x+3}$)  (ĐKXĐ: x $\neq$ ±$\frac{3}{2}$ , x $\neq$ -3)

       = $\frac{3x(2x-3)-4(2x+3)-4x^{2}+23x+12}{4x^{2}-9}$ . $\frac{2x+3}{x+3}$ 

       = $\frac{2x^{2}+6x}{(2x-3)(x+3)}$ 

       = $\frac{2x}{2x-3}$ 

b, $2x^{2}$ + 7x +3=0 ⇔ 2$x^{2}$ + 6x + x + 3=0 ⇔ (x+3)(2x+1)=0 ⇔\(\left[ \begin{array}{l}x+3=0\\2x+1=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-3(loại)\\x=\frac{-1}{2}(TM)\end{array} \right.\)  

 Thay x=$\frac{-1}{2}$ vào B ta có:

B= $\frac{1}{4}$ 

c, Để B ∈ Z thì $\frac{2x}{2x-3}$ ∈ Z

⇒ 2x chia hết cho 2x - 3

⇔ 2x-2x+3 chia hết cho 2x-3

⇔ 3 chia hết cho 2x-3

⇔ 2x-3 ∈ Ư(3)= {±1;±3}

⇔ x ∈ {2;1;3;0}

d, |B| < 1 ⇔ \(\left[ \begin{array}{l}0\leq B <1\\0\geq B >-1 \end{array} \right.\)