[tex]\displaystyle\bf\\A=\left\{n\in Z^\bigstar\left|~\frac{5n+19}{2n+1}\in Z\right\}\right\\\\\implies (5n+19)~\vdots~(2n+1)\\\\\implies 2\times(5n+19)~\vdots~(2n+1)\\\\\frac{2\times(5n+19)}{2n+1}=\frac{10n+38}{2n+1}=\\\\=\frac{10n+5+33}{2n+1}=\frac{10n+5}{2n+1}+\frac{33}{2n+1}=5+\frac{33}{2n+1}\\\\5\in Z\\\\\implies \frac{33}{2n+1}\in Z\\\\D_{33}=\{-33;~-11;~-3;~3;~11;~33\}\\\\2n+1=-33;~n=-17\\2n+1=-11;~n=-6\\2n+1=-3;~n=-2\\2n+1=3;~n=1\\2n+1=11;~n=5\\2n+1=33;~n=16\\A=\{-17;-6;-2;1;5;16\}[/tex]
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[tex]\displaystyle\bf\\B=\left\{n\in Z~\left|~\frac{7n+10}{2n-1}\in Z \right\}\right\\\\\implies~(7n+10)~\vdots~(2n-1)\\\implies~2\times(7n+10)~\vdots~(2n-1)\\\\\frac{2(7n+10)}{2n-1}=\frac{14n+20}{2n-1}=\frac{14n-7+27}{2n-1}=\\\\=\frac{14n-7}{2n-1}+\frac{27}{2n-1}=7+\frac{27}{2n-1}\\\\7\in Z\\\implies \frac{27}{2n-1}\in Z\\\\D_{27}=\{-27;~-9;~-1;~1;~9;~27\}\\\\2n-1=-27;~n=-13\\2n-1=-9;~n=-4\\2n-1=-1;~n=0\\2n-1=1;~n=1\\2n-1=9;~n=5\\2n-1=27;~n=14\\A=\{-13;~-4;~0;~1;~5;~14\}\\\\A\bigcap B=\{1;~5\}[/tex]
