Đáp án:
$3)\\ a) \left[\begin{array}{l} x=1\\x=-\dfrac{3}{2}\end{array} \right.\\ b) \left[\begin{array}{l} m=\dfrac{3}{4}\\m=-\dfrac{3}{4}\end{array} \right.$
Giải thích các bước giải:
$3)\\ x^2-2(m-1)x-2m=0(1)\\ a)m=\dfrac{3}{4}; (1) \Leftrightarrow x^2+\dfrac{1}{2}x-\dfrac{3}{2}=0\\ \Leftrightarrow \left[\begin{array}{l} x=1\\x=-\dfrac{3}{2}\end{array} \right.\\ b)\Delta'=(m-1)^2+2m=m^2+1>0 \ \forall \ m\\ Vi-et: x_1+x_2=2(m-1) \Rightarrow x_2=2(m-1)-x_1\\ x_1x_2=-2m\\ \circledast x_1^2+x_1-x_2=5-2m\\ \Leftrightarrow x_1^2+x_1-2(m-1)+x_1-5+2m=0\\ \Leftrightarrow x_1^2+2x_1−3=0\\ \Leftrightarrow \left[\begin{array}{l} x_1=1 \Rightarrow x_2=2m-3\\x_1=-3 \Rightarrow x_2=2m+1\end{array} \right.\\ \Rightarrow \left[\begin{array}{l} x_1x_2=2m-3\\x_1x_2=-3(2m+1)\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} -2m=2m-3\\-2m=-3(2m+1)\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} 4m=3\\4m=-3\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} m=\dfrac{3}{4}\\m=-\dfrac{3}{4}\end{array} \right.$