a ) a) a ) ( 1 ) (1) ( 1 ) a ≠ 0 a\ne0 a ̸ = 0
$⇔
$⇔
b ) ( 1 ) b)(1) b ) ( 1 ) a > 0 a>0 a > 0
⇔ ⇔ ⇔ [ m − 1 > 0 m + 1 > 0 \left[ \begin{array}{l}m-1>0\\m+1>0\end{array} \right. [ m − 1 > 0 m + 1 > 0 [ m − 1 < 0 m + 1 < 0 \left[ \begin{array}{l}m-1<0\\m+1<0\end{array} \right. [ m − 1 < 0 m + 1 < 0
⇔ ⇔ ⇔ [ m > 1 m > − 1 \left[ \begin{array}{l}m>1\\m>-1\end{array} \right. [ m > 1 m > − 1 [ m < 1 m < − 1 \left[ \begin{array}{l}m<1\\m<-1\end{array} \right. [ m < 1 m < − 1
⇔ ⇔ ⇔ [ m > 1 m < − 1 \left[ \begin{array}{l}m>1\\m<-1\end{array} \right. [ m > 1 m < − 1
c ) c) c ) A ( 1 ; 2 ) A(1;2) A ( 1 ; 2 ) ( 1 ) (1) ( 1 )
m − 1 m + 1 + m + 2 = 2 ( m ≠ − 1 ) \dfrac{m-1}{m+1}+m+2=2(m\ne-1) m + 1 m − 1 + m + 2 = 2 ( m ̸ = − 1 )
⇔ m − 1 + m ( m + 1 ) m + 1 = 0 ⇔\dfrac{m-1+m(m+1)}{m+1}=0 ⇔ m + 1 m − 1 + m ( m + 1 ) = 0
⇔ m − 1 + m 2 + m = 0 ⇔m-1+m^2+m=0 ⇔ m − 1 + m 2 + m = 0
⇔ m 2 + 2 m − 1 = 0 ⇔m^2+2m-1=0 ⇔ m 2 + 2 m − 1 = 0
⇔ ( m + 1 ) 2 = 2 ⇔(m+1)^2=2 ⇔ ( m + 1 ) 2 = 2
⇔ ⇔ ⇔ [ m + 1 = 2 m + 1 = − 2 \left[ \begin{array}{l}m+1=\sqrt[]{2}\\m+1=-\sqrt[]{2}\end{array} \right. [ m + 1 = 2 m + 1 = − 2
⇔ ⇔ ⇔ [ m = 2 − 1 m = − 1 − 2 \left[ \begin{array}{l}m=\sqrt[]{2}-1\\m=-1-\sqrt[]{2}\end{array} \right. [ m = 2 − 1 m = − 1 − 2
