Đề khá hay đấy! Nhưng lần sau đừng viết sai đề nx!
a) ĐK: \(x>4\)
b) \(P=\dfrac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\dfrac{8}{x}+\dfrac{16}{x^2}}}\)
= \(\dfrac{\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}}{\sqrt{1-2.\dfrac{4}{x}+\left(\dfrac{4}{x}\right)^2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\sqrt{\left(1-\dfrac{4}{x}\right)^2}}\)
= \(\dfrac{\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|}{\left|1-\dfrac{4}{x}\right|}\)
= \(\dfrac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\dfrac{4}{x}}\) = \(\left[{}\begin{matrix}\dfrac{2x\sqrt{x-4}}{x-4}khix\ge8\\\dfrac{4x}{x-4}khi4< x< 8\end{matrix}\right.\)
Xét \(P=\dfrac{2x}{\sqrt{x-4}}\left(x\ge8\right)\) thì:
Để \(P\in Z\) khi \(\dfrac{2x-8+8}{\sqrt{x-4}}\in Z\)
<=> \(2.\left(\sqrt{x-4}\right)+\dfrac{8}{\sqrt{x-4}}\in Z\)
<=> \(\left\{{}\begin{matrix}\sqrt{x-4}\in Z^+\\\sqrt{x-4}\inƯ\left(8\right)\end{matrix}\right.\)
Mà \(x\ge8\) => \(\left[{}\begin{matrix}\sqrt{x-4}=2\\\sqrt{x-4}=4\\\sqrt{x-4}=8\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=8\\x=20\\x=68\end{matrix}\right.\)
Xét \(P=\dfrac{4x}{x-4}\left(4< x< 8\right)\) thì:
Để \(P\in Z\) khi \(\dfrac{4x-16+16}{x-4}\in Z\) <=> \(4+\dfrac{16}{x-4}\in Z\)
=> \(x-4\inƯ\left(16\right)\) mà \(0< x-4< 4\)
=> \(x-4=2\) => \(x=6\)
Vậy \(x\in\left\{6;8;20;68\right\}\) thì \(P\in Z\)
P/s: Vì bài này dài nên mk lm khá tắt, ko hiểu cứ hỏi!
