Explicație pas cu pas:
2. 15/(2x-1)€Z <=> 2x-1€D15 in Z =>
2x-1€{-15,-5,-3,-1,1,3,5,15} =>
2x€{-14,-4,-2,0,2,4,6,16} =>
x€{-7,-2,-1,0,1,2,3,4}
(4x+23)/(2x+1)€Z <=>
2x+1|4x+23
2x+1|2x+1=> 2x+1|4x+2
=> 2x+1|4x+23-4x-2=>2x+1|21 =>
2x+1€D21 in Z =>
2x+1€{-21,-7,-3,-1,1,3,7,21} =>
2x€{-22,-8,-4,-2,0,2,6,20} =>
x€{-11,-4,-2,-1,0,1,3,10}
3.V(7+4V3)=V(4+4V3+3)=V(2+V3)²=|2+V3|=2+V3
V(52-14V3)=V(49-14V3+3)=V(7-V3)²=|7-V3|=7-V3
=> (2+V3+7-V3)/(2x-1)€Z => 9/(2x+1)€Z <=> 2x+1€D9 in Z =>
2x+1€{-9,-3,-1,1,3,9} =>
2x€{-10,-4,-2,0,2,8} =>
x€{-5,-2,-1,0,1,4}
4.V(0,a(b)+0,b(a)) €Q <=> 0,a(b)+0,b(a) patrat perfect
=> (|ab -a)/90 + (|ba -b)/90 p.p =>
(10a+b-a)/90+(10b+a-b)/90 p.p =>
(9a+b)/90 + (9b+a)/90 p.p =>
(10a+10b)/90 p.p =>
10(a+b)/90 p.p =>
(a+b)/9 p.p <=>
a+b=0 (imposibil)
a+b=1(imposibil pentru ca a,b cifre dif de 0)
a+b=4 (adevarat)
a+b=9(adevarat)
a+b=16(adevarat)
De aici scoti cazuri pentru a si b si alea sunt cifrele cautate(vezi sa nu uiti ca a<b)
a+b=4 => a=1 b=3 (singura varianta pentru ca mai avem a=2 b=2 imposibil de la a<b si a=3 si b=1 imposibil)
a+b=9 => a=1 b=8,a=2 b=7, a=3 b=6, a=4 b=5
a+b=16=> a=7 b=9(singura varianat pentru ca a si b sunt cifre)
