Bài 3:
a, $A=\frac{m+1}{m-2}-$ $\frac{1}{m}$
⇒$A=\frac{m(m+1)-(m-2)}{m(m-2)}$
⇒$A=\frac{m^2+m-m+2}{m(m+2)}$
⇒$A=\frac{m^2+2}{m(m+2)}$
$B=\frac{1}{m}+$ $\frac{2+m}{m-2}$
⇒$B=\frac{m-2+m(2+m)}{m(m-2)}$
⇒$B=\frac{m-2+2m+m^2}{m(m-2)}$
⇒$B=\frac{m^2+3m-2}{m(m-2)}$
b, Ta có: $A=B$
⇔$\frac{m^2+2}{m(m+2)}=\frac{m^2+3m-2}{m(m-2)}$
⇔$\frac{(m-2)(m^2+2)-(m+2)(m^2+3m-2)}{m(m-2)(m+2)}=0$
⇔$(m-2)(m^2+2)-(m+2)(m^2+3m-2)=0$
⇔$-m(7m+2)=0$
⇔\(\left[ \begin{array}{l}-m=0\\7m+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}m=0\\m=-2/7\end{array} \right.\)
Vậy $S=${$0;-2/7$}
c, Ta có: $A=1$
⇔$\frac{m^2+2}{m(m+2)}=1$
⇔$\frac{m^2+2-(m^2+2m)}{m(m+2)}=0$
⇔$m^2+2-m^2-2m=0$
⇔$2-2m=0$
⇔$2m=2$
⇔$m=1$
Vậy $S=${$1$}
d, Ta có: $A+B=0$
⇔$\frac{m^2+2}{m(m+2)}+\frac{m^2+3m-2}{m(m-2)}=0$
⇔$\frac{(m-2)(m^2+2)+((m+2)(m^2+3m-2)}{m(m-2)(m+2)}=0$
⇔$(m-2)(m^2+2)+(m-2)(m^22+3m-2)=0$
⇔$2m^3+6m-8=0$
⇔$$
