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: $A+B=0$

⇔$\frac{m^2+2}{m(m+2)}+\frac{m^2+3m-2}{m(m-2)}=0$

⇔$\frac{(m-2)(m^2+2)+((m+2)(m^2+3m-2)}{m(m-2)(m+2)}=0$

⇔$(m-2)(m^2+2)+(m-2)(m^22+3m-2)=0$

⇔$2m^3+6m-8=0$

⇔$$

Đáp án:

CHÚC BẠN HỌC TỐT!!! 

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