Giải thích các bước giải:

a.ĐKXĐ: $x\ne\pm\dfrac12$

Ta có:

$B=(\dfrac2{2x+1}-\dfrac3{1-4x^2}-\dfrac2{2x-1}):\dfrac{4x^2+1}{4x^2-1}$

$\to B=(\dfrac2{2x+1}+\dfrac3{4x^2-1}-\dfrac2{2x-1}):\dfrac{4x^2+1}{4x^2-1}$

$\to B=(\dfrac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{3}{\left(2x+1\right)\left(2x-1\right)}-\dfrac{2\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}):\dfrac{4x^2+1}{4x^2-1}$

$\to B=(\dfrac{2\left(2x-1\right)+3-2\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}):\dfrac{4x^2+1}{4x^2-1}$

$\to B=\dfrac{-1}{\left(2x+1\right)\left(2x-1\right)}:\dfrac{4x^2+1}{(2x-1)(2x+1)}$

$\to B=-\dfrac1{4x^2+1}$

b.Khi $x=-\dfrac23$

$\to B=-\dfrac1{4\cdot (-\dfrac23)^2+1}=-\dfrac9{25}$

c.Vì $4x^2+1>0,\quad\forall x$

$\to B=\dfrac{-1}{4x^2+1}<0,\quad\forall x\ne\pm\dfrac12$

d.Ta có:

$B=-\dfrac1{4x^2+1}\ge -\dfrac1{4\cdot 0+1}=-1$

$\to GTNN_B=-1\to x=0$