(2x-1)²=25⇒2x-1=+sau -√25⇒2x-1=+sau-5⇒2x-1=5⇒2x=5+1⇒2x=6⇒x=6/2=3
si 2x-1=-5⇒2x=-5+1⇒2x=-4⇒x=-4/2=-2 avem solutiile -2 si 3
Ι2x-1Ι≤3rezulta -3≤2x-1≤3 adunam 1 la toata relatia rezulta -2≤2x≤4 ,impartim prin 2 rezulta -1≤x≤2 rezulta x∈ multimii -1,0,1,2 daca lucram in Zsau x∈multimii 0,1,2 daca lucram in N
