varianta 1.mai simplu se vede pe atasament unde am scris fiecare expresie si fiecare ecuatie
varianta 2 am corectat solutia initiala ; cam lunga
|x+1|=0 pt x=-1
verificam pt x=-1
2^0=|1/2-1|+1/2+1
1=1/2=1/2+1. fals, -1nu este solutie
|x+1|=x+1 pt x>-1
2^ (x+1)=|2^x-1|+2^x+1
pt x∈(-1;0), |2^x-1|=1-2^x, deci
2^(x+1)=1-2^x+2^x+1
2^(x+1)=2 adica x=0 dar care∉(-1;0)
dar vom verifica pt x=0
2=1+1 adevarat deci x=0 este solutie
pt x∈(0;1), |2^x-1|=2^x-1
2^(x+1)=2^x-1+2^x+1
2^(x+1)=2*2^x=2^(x+1) identitate, valabil∀x(0;1)
pt x=1
2²=1+2+1
4=4 adevarat
deci x∈(0;1) ∪{0}∪{1}=[0;1]
pt x>1
2^(x+1)= 2^x-1+2^x+1
2^(x+1)=2^x+2^x=2*2^x=2^(x+1)
identitate, valabil ∀x∈(1;∞)
deci [0;1]∪(1;∞)=[0;∞)
pt x<-1. |x+1|=-x-1
si|2^x-1|=1-2^x
atunci
2 ^ (-x-1)=1-2^x+2^x+1
2^(-x-1)=2=2^1
-x-1=1
-x=2
x=-2∈(-∞;-1)
verificare
2=|-3/4|+1/4+1 adevarat
deci solutie finala
x∈{-2}∪[0;∞)
