a:
ĐKXĐ: a>4
\(P=\dfrac{\left|\sqrt{a-4}+2\right|+\left|\sqrt{a-4}-2\right|}{\sqrt{\dfrac{a^2-8a+16}{a^2}}}\)
\(=\dfrac{\sqrt{a-4}+2+\left|\sqrt{a-4}-2\right|}{\dfrac{\left|a-4\right|}{a}}\)
\(=\left(\sqrt{a-4}+2+\left|\sqrt{a-4}-2\right|\right)\cdot\dfrac{a}{a-4}\)
TH1: a>=8
\(A=\left(\sqrt{a-4}+2+\sqrt{a-4}-2\right)\cdot\dfrac{a}{a-4}=2\sqrt{a-4}\cdot\dfrac{a}{a-4}\)
\(=\dfrac{2a}{\sqrt{a-4}}\)
TH2: 4<a<8
\(A=\left(\sqrt{a-4}+2+2-\sqrt{a-4}\right)\cdot\dfrac{a}{a-4}=\dfrac{4a}{a-4}\)
b: TH1: a>=8
Để P là số nguyên thì \(2a⋮\sqrt{a-4}\)
=>\(2a-8+8⋮\sqrt{a-4}\)
=>\(\sqrt{a-4}\in\left\{1;2;4;8\right\}\)
=>\(a\in\left\{8;20;68\right\}\)
TH2: 4<a<8
Để P là số nguyên thì 4a-16+16 chia hết cho a-4
=>\(a-4\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
hay \(a\in\left\{5;6\right\}\)
