1.a) punem conditia 3x+5 >0
facem tabel de semne pt fct de gr I
x -∞ [tex] \frac{-5}{3} [/tex] +∞
3x+5 -------------------0++++++++++++++++
3x+5>0 pentru x∈( [tex] \frac{-5}{3} [/tex] ,+∞) ⇒ ca D= ( [tex] \frac{-5}{3} [/tex] ,+∞)
b)x∈R⇒D=R
2.a) [tex] \sqrt[3]{2x+3} [/tex] =-3 |∛ punem cond. x∈R
2x+3=∛-3
2x+3=-27
2x=-30
x=-15 ∈R
b)1+[tex] \sqrt{7-x} [/tex] =x
[tex] \sqrt{7-x} [/tex]= x-1 |[tex] \sqrt{} [/tex] p.c. 7-x≥ 0 si x-1 ≥0 7-x= (x-1)²
7-x= x²-2x+1
7-x-x²+2x-1=0
-x²+x+6=0
x²-x-6=0
Δ=b²-4ac
Δ=1+24
Δ=25>0
x1,2=(-b+-√Δ)/2a
x1,2=(1+-5)/2 ⇒ x1=3
x2=-2
x∈{-2,3}
Verificam conditiile
Daca x=-2 atunci 7-(-2)≥0 (A)
(-2)-1≥0(F)
Deci, x=-2 nu este solutie
Daca x=3 atunci 7-3≥0 (A)
3-1≥0 (a)
Deci, x=3 este singura solutie a ecuatiei
c)[tex] \sqrt{6-4x-x2} [/tex] =x+4 |√ p.c 6-4x-x²≥0 si x+4≥0
6-4x-x²=(x+4)²
6-4x-x²=x²+8x+16
6-4x-x²-x²-8x-16=0
-2x²-12x-10=0
2x²+12x+10=0
x²+6x+5=0
Δ=36-25
Δ=11
x1=-6+√11/4
x2=-6-√11/4
Verif conditiile
Daca x=-6+11/4 atunci 6-4*-6+11/4 - (-6+11/4)²≥0 (A)
-6+11/4+4≥0 (A)
Daca x=-6-√11/4 atunci 6-4*-6-4*-6+11/4 - (-6-11/4)²≥0 (A)
-6-11/4+4≥0 (F)
Deci, x=-6+11/4 este solutie
