解:deta=m^2-4m(5-m)=0m1=0舍,m2=4m=4:x1=x2=-1/2
mx^2+mx+5-m=0m(x^2+x+(5-m)/m)=0所以（5-m）/m=1/4m=4
mx^2+mx+(5-m)=0m^2-4m*(5-m)=0m^2-4m=0m=4 (m=0舍弃)代入原方程,4x^2+4x+1=0(2x+1)^2=0x=-1/2
