xác định m để hpt mx-2y=m^2-m+6;(m+1)x-2y=m^2+7 có no(x;y)mà điểm (x;y) thuộc đường thẳng 2x-y+3=0 27/10/2021 Bởi Parker xác định m để hpt mx-2y=m^2-m+6;(m+1)x-2y=m^2+7 có no(x;y)mà điểm (x;y) thuộc đường thẳng 2x-y+3=0
$\left\{\begin{array}{l} mx-2y=m^2-m+6\\ (m+1)x-2y=m^2+7\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx-2y=m^2-m+6\\ (m+1)x-2y-(mx-2y)=m^2+7-(m^2-m+6)\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx-2y=m^2-m+6\\x=m+1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=m+1\\y=\dfrac{mx-m^2+m-6}{2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=m+1\\y=\dfrac{m(m+1)-m^2+m-6}{2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=m+1\\y=m-3\end{array} \right.\\ (x;y) \in (d):2x-y+3=0\\ \Leftrightarrow 2(m+1)-m+3+3=0\\ \Leftrightarrow m=-8$ Bình luận
$\left\{\begin{array}{l} mx-2y=m^2-m+6\\ (m+1)x-2y=m^2+7\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx-2y=m^2-m+6\\ (m+1)x-2y-(mx-2y)=m^2+7-(m^2-m+6)\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx-2y=m^2-m+6\\x=m+1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=m+1\\y=\dfrac{mx-m^2+m-6}{2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=m+1\\y=\dfrac{m(m+1)-m^2+m-6}{2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=m+1\\y=m-3\end{array} \right.\\ (x;y) \in (d):2x-y+3=0\\ \Leftrightarrow 2(m+1)-m+3+3=0\\ \Leftrightarrow m=-8$