1, CE
x²-6x+25>0 ,∀x∈R pt ca Δ=36-100<0
ramane ca x+5≥0, deci x≥-5
ridicam la patrat ambii membti
x²-6x+25=x²+10x+25
16x=0
x=0∈[-5;∞) solutia convine
2.
C.E.
(x≥-1/5)∩(x≥-7/6)=x≥-1/5
pt continuare, vezi attach
3
C.E. x∈R, radicalul fiind impar
ridicam la puterea a treia fiecare membru
x³-4x² +6x+2= x³-6x²+12x-8
-4x²+6x+2=-6x²+12x-8
-2x²+3x+1=-3x²+6x-4
x²-3x+5=0
Δ=9-20<0, x1,2∉R