a) \(\Delta\)' = \(m^2-m^2+4=4>0\forall m\)
\(\Rightarrow\) pt có 2 nghiệm phân biệt \(\forall m\)
b) ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m\\2x_1-x_2=0\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x_1=2m\\x_1+x_2=2m\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x_1=\dfrac{2m}{3}\\\dfrac{2m}{3}+x_2=2m\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x_1=\dfrac{2m}{3}\\x_2=\dfrac{4m}{3}\end{matrix}\right.\)
ta có : \(x_1x_2=m^2-4\) \(\Leftrightarrow\) \(\dfrac{8m^2}{9}=m^2-4\)
\(\Leftrightarrow\) \(8m^2=9m^2-36\) \(\Leftrightarrow\) \(m^2=36\) \(\Leftrightarrow\) \(m=\pm6\)
vậy \(m=\pm6\) thỏa mảng đk bài toán
c) ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m\\3x_1+2x_2=7\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x_1+2x_2=4m\\3x_1+2x_2=7\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x_1=7-4m\\7-4m+x_2=2m\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x_1=7-4m\\x_2=6m-7\end{matrix}\right.\)
ta có : \(x_1x_2=m^2-4\) \(\Leftrightarrow\) \(\left(7-4m\right)\left(6m-7\right)=m^2-4\)
\(\Leftrightarrow\) \(42m-49-24m^2+28m=m^2-4\)
\(\Leftrightarrow\) \(25m^2-70m+45=0\)
\(\Leftrightarrow\) \(5m^2-14m+9=0\)
giải phương trình ta có : \(\left\{{}\begin{matrix}x=\dfrac{9}{5}\\x=1\end{matrix}\right.\)
vậy : \(x=\dfrac{9}{5};x=1\) thỏa mãng đk bài toán
