$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\frac{1}{x} +\frac{2}{x-1} +\frac{1}{x^{2} -x} \ ( x\#0;x\#1)\\ A=\frac{1}{x} +\frac{2}{x-1} +\frac{1}{x( x-1) \ }\\ A=\frac{x-1+2x+1}{x( x-1)} =\frac{3x}{x( x-1)} =\frac{3}{x-1}\\ b) \ ( \ vì\ 0\ rõ\ nguyên\ x\ nên\ mình\ làm\ tổng\ quát\ ) \ \\ Vì\ A\ nguyên\ nên\ \frac{3}{x-1} \ nguyên\ \\ \rightarrow \ x-1=\frac{3}{a} \ ( a\ thuộc\ Z;a\#0\ ) \ \\ \rightarrow x=\frac{3}{a} +1\ \\ Để\ \frac{3}{a} +1\ \#0\ \\ \rightarrow a\#-3\ \\ \frac{3}{a} +1\#1\ \\ \rightarrow a\#0( \ vừa\ đúng\ dk\ ) \ \ \\ Vậy\ Để\ A\ nguyên\ thì\ x=\frac{3}{a} +1;\ a\#-3;a\#0\ và\ a\ thuộc\ Z\ \\ Câu\ 3:\ \\ x^{2} -2x-3=0\ \\ \rightarrow x^{2} -3x+x-3=0\ \\ x( x-3) +( x-3) =0\ \\ ( x-3)( x+1) =0\ \\ \rightarrow \left[ \begin{array}{l l} x-3=0 & \\ x+1=0\ & \end{array} \right.\rightarrow \left[ \begin{array}{l l} x=3 & \\ x=-1 & \end{array} \right. \ \ \\ \end{array}$
