Răspuns:
a) Conditie x+2≥0 x≥ -2
Ridici ecuatia la patrat
x²+2x+4=(x+2)²
x²+2x+4=x²+4x+4
2x=4x
-2x=0
x=0
___________
4³ˣ⁻²=8ˣ⁺¹
(2)²⁽³ˣ⁻²⁾=2)³⁽ˣ⁺¹⁾
2⁶ˣ⁻⁴=2³ˣ⁺³
6x-4=3x+3
rezolvi ecuatia
-------------------------------
c.log₅(3x-4)=log₅x
3x-4=x
3x-x=4
2x=4
x=2
d.[tex]\sqrt[3]{x^3+x^2-3x+2} =x[/tex]
Ridici ecuatia la a 3-a
x³+x²-3x+2=x³
x²-3x+2=0
Rezolvi ecuatia de grd2
log₂(x+2)-log₂(x+1)=2
DVA
x+2>0
x+1>0
x> -1
log₂(x+2)/(x+1)=2
(x+2)/(x+1)=2²
x+2=4(x+1)
x+2=4x+4
x-4x=4-2
-3x=2
x=[tex]\frac{-2}{3} >-1[/tex]
Explicație pas cu pas:
