Đáp án:
a)
$4n+3\vdots 2n+1\Rightarrow 4n+3-2(2n+1)\vdots2n+1\Rightarrow 1\vdots 2n+1$
$\Rightarrow 2n+1\in Ư(1)=\pm1$
$\Rightarrow n=-1 (loại)$ hoặc $n=0(tm)$
b)
$x+16\vdots x+1\Rightarrow x+16-(x+1)\vdots x+1\Rightarrow 15\vdots x+1$
$\Rightarrow x+1\in Ư(15)={1,3,5,15}$
TH1: $x+1=1 \Rightarrow x=0(tm)$
TH2: $x+1=3\Rightarrow x=2(tm)$
TH3: $x+1=5 \Rightarrow x=4(tm)$
TH4: $x+1=15 \Rightarrow x=14(tm)$
c)
$x+11\vdots x+4\Rightarrow x+11-(x+4)\vdots x+4\Rightarrow 7\vdots x+4$
$\Rightarrow x+4\in Ư(7)={1,7}$
TH1: $x+4=1\Rightarrow x=-3(loại)$
TH2: $x+4=7\Rightarrow x=3(tm)$
