by Tìm tham số $m$ để hệ phương trình sau vô nghiệm: $\left\{ \begin{array}{l} mx+y+m=0 \\ x+my+m=0 \end{array} \right. $
A. $m=-1. $
B. $m=1. $
C. $m=0. $
D. $m\ne 1. $
Hướng dẫn
$\left\{ \begin{array}{l} mx+y+m=0 \\ x+my+m=0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} mx+y=-m \\ x+my=-m \end{array} \right. $ $D=\left| \begin{array}{l} m\,\,\,\,\,\,\,\,\,1 \\ 1\,\,\,\,\,\,\,\,\,\,m \end{array} \right|={{m}^{2}}-1;$ ${{D}_{x}}=\left| \begin{array}{l} -m\,\,\,\,\,\,\,1 \\ m\,\,\,\,\,\,\,\,\,-m \end{array} \right|={{m}^{2}}-m;$ ${{D}_{y}}=\left| \begin{array}{l} m\,\,\,\,\,\,\,\,-m \\ 1\,\,\,\,\,\,\,\,\,\,-m \end{array} \right|=-{{m}^{2}}+m$ Hệ vô nghiệm $\Leftrightarrow \left\{ \begin{array}{l} D=0 \\ \left[ \begin{array}{l} {{D}_{x}}\ne 0 \\ {{D}_{y}}\ne 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {{m}^{2}}-1=0 \\ \left[ \begin{array}{l} {{m}^{2}}-m\ne 0 \\ -{{m}^{2}}+m\ne 0 \end{array} \right. \end{array} \right. \Rightarrow m=-1. $ Chọn đáp án A.
