Question

1.Se dau multimile: A= ( x∈R| |2x-1| | ≤5 ) si B= (x∈R| -3 (mai mic) $\frac{x+1}{2} \leq 1)$ 
Calculati : A (reunit cu ) B , A (intersectat cu) B , A - B, B-A
2. Se considera numarul $a= \frac{2}{1*3} + \frac{2}{3*5} + .... + \frac{2}{49*51}$
Stabiliti daca a∈$( \frac{1}{2} ; 1 )$

Answer 1

I2x-1I≤5
avem 2 cazuri
I caz 
2x-1≤5
2x≤5+1
2x≤6
x≤6/2
x≤3
II caz:
-2x+1≤5
-2x≤5-1
-2x≤4
-x≤4/2
-x≤2      ·(-1)
x≥-2
solutia x∈[-2,3]
deci A={-2,-1,0,1,2,3}

-3<(x+1)/2
-3·2<x+1
-6<x+1
-x<6+1
-x<7     ·(-1)
x>-7

(x+1)/2≤1
x+1≤2
x≤2-1
x≤1
solutia x∈(-7,1]
deci B={-6,-5,-4,-3,-2,-1,0,1}

AUB={-6,-5,-4,-3,-2,-1,0,1,2,3}
AintersB={-2,-1,0,1}
A-B={-2,-1,0,1,2,3}
B-A={-6,-5,-4,-3,-2,-1,0,1}