Răspuns:
7. m∈ R, x²+2mx>1
x²+2mx-1>0<=> Δ<0si a>0 (1>0- ad)
<=> Δ<0
<=> 4m²+4<0 <=> 4(m²+1)<0 <=> m²+1<0 => m∈Ф
8. f: [0,4] -> R, f(x)= -x²+6x-5
a=-1 <0 => Imf= (-∞; -Δ/4a)= (-∞; 4)
