Giải giúp mk vs
a) \(H=\dfrac{9}{\sqrt{n}-5}\)
Để \(H=\dfrac{9}{\sqrt{n}-5}\in Z\) thì \(9⋮\sqrt{n}-5\)
Vì \(9⋮\sqrt{n}-5\Rightarrow\sqrt{n}-5\inƯ\left(9\right)=\left\{1;3;9;-1;-3;-9\right\}\)
\(\Rightarrow\sqrt{n}\in\left\{6;8;14;4;2;-4\right\}\)
Vì n>0 và \(n\ne25\) nên \(n=2\)
Vậy n=2
b) \(P=\dfrac{3n+2}{n-1}\)
Để \(P=\dfrac{3n+2}{n-1}\in Z\) thì \(3n+2⋮n-1\)
\(\Rightarrow3\left(n-1\right)+5⋮n-1\)
Vì \(3\left(n-1\right)⋮n-1\) nên \(5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)=\left\{1;5;-1;-5\right\}\)
\(\Rightarrow n\in\left\{2;6;0;-4\right\}\)
Vì \(n\ne1\) nên \(n\in\left\{2;6;0;-4\right\}\)
Vậy \(n\in\left\{2;6;0;-4\right\}\)
