$\displaystyle \begin{array}{{>{\displaystyle}l}} B=1+\left(\frac{x+1}{x^{3} +1} -\frac{1}{x -x^{2} -1} -\frac{2}{x-1}\right) :\frac{x^{3} -2x^{2}}{x^{3} -x^{2} +x} \ \\ DKXD:\ x\#\pm 1\ ;x\#2\\ B=1+\left(\frac{x+1}{( x+1)\left( x^{2} -x+1\right)} +\frac{1}{x^{2} -x+1} -\frac{2}{x-1}\right) .\frac{x\left( x^{2} -x+1\right)}{x^{2}( x-2)} \ \\ B=1+\frac{x+1+x+1-2\left( x^{3} +1\right)}{( x +1)\left( x^{2} -x+1\right)( x-1)} .\frac{x\left( x^{2} -x+1\right)}{x^{2}( x-2)}\\ B=1+\frac{2\left( x+1-x^{3} -1\right)}{x^{2} -1} .\frac{1}{x( x-2)}\\ B=1+\frac{2\left( x-x^{3}\right)}{\left( x^{2} -1\right)( x-2) x} \ \\ B=1-\frac{2x\left( x^{2} -1\right)}{\left( x^{2} -1\right)( x-2) x} =1-\frac{2}{x-2} =\frac{x-2-2}{x-2} =\frac{x-4}{x-2} \ \\ b) \ \left| x-\frac{3}{4}\right| =\frac{5}{4} \ \\ \rightarrow \ \left[ \begin{array}{l l} x-\frac{3}{4} =\frac{5}{4}\left( x\geqslant \frac{3}{4}\right) & \\ -x+\frac{3}{4} =\frac{5}{4}\left( x< \frac{3}{4}\right) & \end{array} \right.\rightarrow \left[ \begin{array}{l l} x=\frac{8}{4} =2( ktm\ DKXD\ ) & \\ x=\frac{-2}{4} =\frac{-1}{2} \ ( tm\ ) & \end{array} \right.\\ B=\frac{-\frac{1}{2} -4}{-\frac{1}{2} -2} =\frac{9}{5} \ \ \\ Vậy\ khi\ x=\frac{-1}{2} \ \ thì\ B=\frac{9}{5} \ \\ c)\frac{x-4}{x-2} =3\ \\ DK:x\#\pm 1;x\#2\ \\ \rightarrow \frac{x-4-3( x-2)}{x-2} =0\ \\ -.x-4-3x+6=0\ \\ \rightarrow -2x+2=0\ \\ \rightarrow x=1\ ( ktm) \ \\ d) \ \frac{x-4}{x-2} =1-\frac{2}{x-2} \ \\ Để\ B\ nguyên\ thì\ \frac{2}{x-2} \ nguyên\ \\ Vì\ x\ nguyên\ nên\ x\ thuộc\ Ư( 2) \ \\ \rightarrow x-2=1\ \rightarrow x=3\ \\ x-2=-1\rightarrow x=1( ktm) \ \\ x-2=2\ \rightarrow x=4\ \\ x-2=-2\ \rightarrow x=0\ \\ Vậy\ để\ B\ nguyên\ thì\ x=\{3;4;0\} \ \\ e) \ Để\ B< 0\ \\ \rightarrow \frac{x-4}{x-2} \ < 0\ \\ \rightarrow \left[ \begin{array}{l l} \begin{cases} x-4< 0 & \\ x-2 >0 & \end{cases} & -.\left[ \begin{array}{l l} \begin{cases} x< 4 & \\ x< 2 & \end{cases} & \\ \begin{cases} x >4 & \\ x< 2\ & \end{cases} & \end{array} \right.\\ \begin{cases} x-4 >0 & \\ x-2< 0 & \end{cases} & \end{array} \right.\\ \rightarrow 2< x< 4\ \\ Để\ B\ < 0\ thì\ 2< x< 4\ \\ B >2\ \\ \rightarrow B-2 >0\ \\ \rightarrow \frac{x-4}{x-2} -2 >0\ \\ \rightarrow \frac{x-4-2( x-2)}{x-2} >0\ \\ \rightarrow \frac{-x}{x-2} >0\ \rightarrow \frac{x}{x-2} < 0\ \\ \left[ \begin{array}{l l} \begin{cases} x< 0 & \\ x-2 >0 & \end{cases} & \\ \begin{cases} x >0 & \\ x-2< 0 & \end{cases} & \end{array} \right.\rightarrow \left[ \begin{array}{l l} \begin{cases} x< 0 & \\ x >2 & \end{cases} & \\ \begin{cases} x >0 & \\ x< 2 & \end{cases} & \end{array} \right.\\ \rightarrow 0< x< 2\\ \\ \\ \end{array}$