[tex]\displaystyle\bf\\a)\\\frac{a}{7-4\sqrt{3}}+\frac{b}{7+4\sqrt{3}}=14+16\sqrt{3}\\\\\\\frac{a\Big(7+4\sqrt{3}\Big)+b\Big(7-4\sqrt{3}\Big)}{\Big(7-4\sqrt{3}\Big)\Big(7+4\sqrt{3}\Big)}=14+16\sqrt{3}\\\\\\\frac{7a+4a\sqrt{3}+7b-4b\sqrt{3}}{7^2-\Big(4\sqrt{3}\Big)^2}=14+16\sqrt{3}\\\\\\\frac{7a+7b+4a\sqrt{3}-4b\sqrt{3}}{49-48}=14+16\sqrt{3}[/tex]
[tex]\displaystyle\bf\\7(a+b)+4\sqrt{3}(a-b)=14+16\sqrt{3}\\\\7(a+b)=14\implies a+b=\frac{14}{7}=\boxed{\bf2}\\\\4\sqrt{3}(a-b)=16\sqrt{3}\implies a-b=\frac{16\sqrt{3}}{4\sqrt{3}}=\boxed{\bf4}\\\\\\a+b=2\\a-b=4\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_~~Adunam~ecuatiile\\2a~~/~=6\\a=6:2\\\boxed{\bf~a=3}\\a+b=2\\b=2-a\\b=2-3\\\boxed{\bf~b=-1}\\\\\\\frac{5a+b}{2a-b}=\frac{5\times3+(-1)}{2\times3-(-1)}=\frac{15-1}{6+1}=\frac{14}{7}=\boxed{\bf2\in N}[/tex]
