2 a) fie∛x=y
ecuatia devine
2y+3y²=5y³
y(2+3y-5y²)=0
y1=0 x1=0³=0
-5y²+3y+2=0
5y²-3y-2=0
3y²-3y+2y²-2=0
3y(y-1) +2(y-1)(y+1)=0
(y-1) (3y+2y+2)=0
(y-1)(5y+2)=0
y2=1 x2=(1)³=-1
y3=-2/5 x3=(-2/5)³=-8/125
b) 2^x+3^x+4^x=3*9^x=9^x+9*x+9^x
ptx>0, 2^x<9^x
3^x<9^x
4^x<9^x
pt x<0 e invers 2^x>9^x
3^x>9^x
4^x>9^x
doar pt x=0 a vem2^x= 3^x=4^x=9^x=1
deci singura solutie x=0
verificare 1+1+1=3
c) 27-3^(1/x)>0
3³>3^(1/x)
baza 3>1 functia 3^x crescatoare
3>1/x
3/1>1/x
1/3<x
x>1/3
x=1/2 aparine domeniului si verifica ecuatia
(1+1) lg3 +lg2=lg(27-9)
lg9+lg2=lg18
cum demonstram ca e unica??
1/2x descrescatoare
(1+1/2x) *lg3 descrescatoare
lg2 constanta
3^(1/x) descrescvatoare
-3^(1/x) crescatoare
27-3^(1/x) crescatoare
lg (27-3^(1/x) ) crescatoare
deci membrul stang descrescator, membru drept crescator solutia x=1/2 este unica
