Question

已知函数fx=x-2/ax+1 若mx=ax+fx 试问方程mx=0有没有负根

Answer 1

f(x)=(x-2)/(ax+1)

m(x)=ax+f(x)=(a^2x^2+ax+x-2)/(ax+1)

=[a^2x^2+(a+1)x-2]/(ax+1)

∵△>0 -2/a^2