(x;y) là nghiệm của hệ phương trình \(\begin{equation} \left\{\begin{array}{l} 2 x+3 y=\frac{7}{2}-m \\ 4 x-y=5 m \end{array}\right. \end{equation}\). Tìm m thỏa \(\begin{equation}...

* Hướng dẫn giải

 \(\begin{equation} \begin{aligned} &\text { Ta có }\left\{\begin{array}{l} 2 x+3 y=\frac{7}{2}-m \\ 4 x-y=5 m \end{array} \Leftrightarrow\left\{\begin{array}{l} 4 x+6 y=7-2 m \\ 4 x-y=5 m \end{array} \Leftrightarrow\left\{\begin{array}{l} 7 y=7-7 m \\ 4 x-y=5 m \end{array}\right.\right.\right.\\ &\Leftrightarrow\left\{\begin{array}{l} y=1-m \\ 4 x-(1-m)=5 m \end{array} \Leftrightarrow\left\{\begin{array}{l} y=1-m \\ x=4 m+14 \end{array}\right.\right.\\ &\text { Thay vào } x^{2}+y^{2}=\frac{25}{16} \text { ta có } x^{2}+y^{2}=\frac{25}{16} \Leftrightarrow\left(\frac{4 m+1}{4}\right)^{2}+(1-m)^{2}=\frac{25}{16}\\ &\Leftrightarrow 16 m^{2}+8 m+1+16 m^{2}-32 m+16=25\\ &\Leftrightarrow 32 m^{2}-24 m-8=0 \Leftrightarrow 4 m^{2}-3 m-1=0\\ &\Leftrightarrow 4 m^{2}-4 m+m-1=0 \Leftrightarrow(4 m+1)(m-1)=0 \Leftrightarrow\left[\begin{array}{l} m=1 \\ m=-\frac{1}{4} \end{array}\right. \end{aligned} \end{equation}\)