Question

Exercitiul 1 punctul a

Answer 1

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;∞)

Answer 2

[tex]\text{Avem de analizat trei cazuri,dar mai intai sa explicitam modulele:}\\ |x+1|=\left \{ {{x+1; x\geq -1} \atop {-x-1;x\ \textless \ -1}} \right. ;|2^x-1|= \left \{ {{2^x-1, x\geq 0} \atop {1-2^x, x\ \textless \ 0}} \right. \\ \text{Daca facem tabel de semne vom avea:}\\ x~~~~~~|-\infty~~~~~~~~~~-1~~~~~~~~~~0~~~~~~~~~+\infty\\ x+1~| ~~~~ -x-1~~~~~ 0 ~~~~~~~~ x+1 \\ 2^x-1|~~~~~~~1-2^x~~~~~~~~~~~~~~~~0~~~~~~~~2^x-1\\ i)Pt. x\in(-\infty;-1):\\ 2^{-x-1}=1-2^x+2^x+1\\ 2^{-x}=4\Rightarrow x=-2(convine)\\ ii)Pt. x\in [-1,0):[/tex]
[tex]2^{x+1}=1-2^x+2^x+1\\ 2^{x+1}=2\\ x+1=1\Rightarrow x=0(nu\ convine)\\ iii)Pt.x\in [0;+\infty):\\ 2^{x+1}=2^x-1+2^x+1\\ 2^{x+1}=2\cdot 2^x\\ 2^{x+1}=2^{x+1}(A)\Rightarrow x\in [0,+\infty)\\ S=[0;\infty)\cup \{-2\}[/tex]