Đáp án+Giải thích các bước giải:

$x^2-5mx-4m=0$

$1)$Phương trình có nghiệm

$\Rightarrow \Delta \ge 0\\ \Leftrightarrow (-5m)^2+16m \ge 0\\ \Leftrightarrow 25m^2+16m \ge 0\\ \Leftrightarrow m(25m+16) \ge 0\\ \Leftrightarrow \left[\begin{array}{l} m \ge 0 \\ m \le -\dfrac{16}{25}\end{array} \right.\\ 2)Vi-et: x_1+x_2=5m\\ x_1x_2=-4m\\ a)x_1^2+5mx_2-4m\\ =x_1^2+(x_1+x_2)x_2+x_1x_2\\ =x_1^2+2x_1x_2+x_2^2\\ =(x_1+x_2)^2\\ =(5m)^2\\ =25m^2 \ge 0\\ b) x_1+x_2=5m \Rightarrow x_1=5m-x_2\\ x_1^2+4x_2^2=5x_1x_2\\ \Leftrightarrow x_1^2+4x_2^2-5x_1x_2=0\\ \Leftrightarrow x_1^2-x_1x_2+4x_2^2-4x_1x_2=0\\ \Leftrightarrow x_1(x_1-x_2)-4x_2(x_1-x_2)=0\\ \Leftrightarrow (x_1-4x_2)(x_1-x_2)=0\\ \Leftrightarrow \left[\begin{array}{l} x_1=4x_2\\x_1=x_2\end{array} \right.\\ \circledast x_1=4x_2\\ x_1+x_2=5m\\ \Leftrightarrow 4x_2+x_2=5m\\ \Leftrightarrow 5x_2=5m\\ \Leftrightarrow x_2=m\\ \Rightarrow x_1=4m\\ x_1x_2=-4m\\ \Leftrightarrow 4m^2=-4m\\ \Leftrightarrow 4m^2+4m=0\\ \Leftrightarrow 4m(m+1)=0\\ \Leftrightarrow m=0(TM); m=-1(TM)\\ \circledast x_1=x_2\\ x_1+x_2=5m\\ \Rightarrow x_1=x_2=\dfrac{5}{2}m\\ x_1x_2=-4m\\ \Leftrightarrow \dfrac{25}{4}m^2=-4m\\ \Leftrightarrow \dfrac{25}{4}m^2+4m=0\\ \Leftrightarrow 25m^2+16m=0\\ \Leftrightarrow m(25m+16)=0\\ \Leftrightarrow m=0(TM); m=-\dfrac{16}{25}(TM)\\ c)x_1<4<x_2\\ \Leftrightarrow (x_1-4)(x_2-4) <0\\ \Leftrightarrow x_1x_2-4(x_1+x_2)+16<0\\ \Leftrightarrow  -4m-4.5m+16<0\\ \Leftrightarrow  -24m+16<0\\ \Leftrightarrow  24m>16\\ \Leftrightarrow m>\dfrac{2}{3}.$