Question

Am nevoie de rezolvari la exercitile incercuite cu rosu.Multumesc!<3

Answer 1

E=21/2x+3=>

2x+3I21
"I"-Divide
2x+3I21
2x+3∈{1,-1,21,-21,3,-3,7,-7}/-3
2x∈{-2,-4,19,-24,0,-6,4,-10}/:2
x∈{-1,-2,-12,0,-3,2,-5}

F=2x+5/x+1
x+1divide2x+5
x+1dividex+1/*2
=>x+1divide2x+5
 x+1divide2x+2
x+1divide3[am scazut din 2x+5 pe 2x+2]
x+1∈{1,3}deoarece x∈N
x∈{0,2}

Answer 2

21/(2x+3) determinam divizorii lui 21={-21,-7,-3,-1,1,3,7,21}
2x+3=-21 ,x=-12
2x+3=-7,x=-5
2x+3=-3,x=-3
2x+3=-1,x=-2
2x+3=1,x=-1
2x+3=3,x=0
2x+3=7 ,x=2
2x+3=21,x=9
x={-12,-5,-3,-2,-1,0,2,9}

(2x+5)/(x+1)=[(2x+2)+3]/(x+1)
avem
2+3/(x+1)
cautam divizorii lui 3={-3,-1,1,3}
x+1=-3,x=-4∉N
x+1=-1,x=-2∉N
x+1=1,x=0
x+1=3,x=2
x={0,2}

(3x+11)/(x+2)=[(3x+6)+5]/(x+2)=3+5/(x+2)
cautam divizorii lui 5={-5,-1,1,5}
x+2=-5,x=-7
x+2=-1,x=-3
x+2=1,x=1
x+2=5,x=3
x={-7,-3,1,3}

(3x+9)/(2x-3)
aici ceva este in neregula 
[3/2(2x-3)+9/2+9]/(2x-3)
3/2+ 27/2(2x+3)
trebuie determinati divizorii lui 27={-27,-9,-3,-1,1,3,9,27}
2x+3=-27
x=-12
2x+3=-3,x=-3
.................x= solutia data in carte