1.
(3-9/4)²·4/3=(x+2)/4
(3/4)²·4/3=(x+2)/4
3/4=(x+2)/4
4x+8=12
4x=4 =>x=1
2.
17/(2x+5)∈R
2x+5≠0 =>2x≠(-5) =>x≠(-5)/2
3.
Daca ΔNBA echilateral =>m(∡B)=60°
In Δdr.AMB avem m(∡MAB)=30°(180-60-90)
In Δdr.AMN avem m(∡NAM)=30°(180-60-90)
=>ΔAMB≡ΔAMN =>MN=MB=3cm
AN=AB=NB (Δ echilat.)
=>AB=6cm
In ΔBAC folosim relatia trigonometrica
cosCBA=cateta alaturata/ipotenuza
Notam segmentul CN=x
CB=6+x
cos∡CBA=AB/BC
cos60°=6/(6+x)
1/2=6/(6+x)
6+x=12
x=6 =>BC=12cm
=>AB+BC=18cm