Cho đường thẳng (d) y= (1-m / m+2) -x + (1-m). (m+2) â) Tìm m để (d) vuong goc (d') y= 1/4.x+1 b) Tìm m để (d) là hàm số đồng biến

$\begin{array}{l} \left( d \right):\,\,\,y = \frac{{1 - m}}{{m + 2}}x + \left( {1 - m} \right)\left( {m + 2} \right)\\ a)\,\,\,d \bot d':\,\,\,y = \frac{1}{4}x + 1\\ \Rightarrow \left\{ \begin{array}{l} \frac{{1 - m}}{{m + 2}}.\frac{1}{4} = - 1\\ m + 2 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 1 - m = - 4m - 8\\ m \ne - 2 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 3m = - 9\\ m \ne - 2 \end{array} \right. \Leftrightarrow m = - 3.\\ b)\,\,Hs\,\,\,DB \Leftrightarrow \frac{{1 - m}}{{m + 2}} > 0\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 1 - m > 0\\ m + 2 > 0 \end{array} \right.\\ \left\{ \begin{array}{l} 1 - m < 0\\ m + 2 < 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} m < 1\\ m > - 2 \end{array} \right.\\ \left\{ \begin{array}{l} m > 1\\ m < - 2 \end{array} \right. \end{array} \right. \Leftrightarrow - 2 < m < 1. \end{array}$