Page 1 of 1

Ecuatie

Posted: Sat Feb 20, 2010 3:30 pm
by Al3xx
Rezolvati in \( R \) ecuatia :

\( |1-x|\ +\ x|x|\ +\ |1+x|\ =\ 3. \)

Posted: Sat Feb 20, 2010 9:41 pm
by Mateescu Constantin
Distingem cazurile :

\( 1.\ x\ \in\ \(-\infty\ ,\ -1\] \) si ecuatia devine : \( -1+x+x(-x)-1-x=3\ \Longleftrightarrow\ x^2=-5 \) , deci nu avem solutii reale .

\( 2.\ x\ \in\ \(-1\ ,\ 0\] \) si ecuatia devine : \( -1+x+x(-x)+1+x=3\ \Longleftrightarrow\ x^2-2x+3=0 \) , fara solutii reale intrucat \( \Delta\ <\ 0 \) .

\( 3.\ x\ \in\ \(0\ ,\ 1\] \) si ecuatia devine : \( -1+x+x^2+1+x=3\ \Longleftrightarrow\ x^2+2x-3=0\ \Longrightarrow\ x=1 \) este solutie .

\( 4.\ x\ \in\ \(1\ ,\ +\infty\) \) si ecuatia devine : \( 1-x+x^2+1+x=3\ \Longleftrightarrow\ x^2=1 \) , ecuatie fara solutii deoarece \( x\ >\ 1 \) .

Prin urmare unica solutie a ecuatiei este \( x=1 \) .