Question

cat mai repede!!!!!​

Answer 1

Răspuns:

36.   (2n-1) | (4n+5)

dar  (2n-1) | (2n-1)

⇔ (2n-1) | (4n+5)

    (2n-1) | 2x(2n-1)

Atunci 2n-1 divide si diferenta

⇔ (2n-1) | (4n+5-4n+2)

⇔ (2n-1) | 7⇒ 2n-1∈D7

2n-1∈{-7,-1,1,7}

2n∈{-6,0,2,8}

⇒n∈{-3,0,1,4}

 dar  n∈Ν

Din cele 2 ecuatii⇒ n∈{0,1,4}

Analog la celelalte