\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne\pm2\end{cases}}\)

\(P=\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{2}{\sqrt{x}+2}-\frac{4\sqrt{x}}{x-4}\)

\(\Leftrightarrow P=\frac{x+2\sqrt{x}-2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Leftrightarrow P=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Leftrightarrow P=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

Để P là số nguyên \(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+2}\)là số nguyên

\(\Leftrightarrow\sqrt{x}-2⋮\sqrt{x}+2\)

\(\Leftrightarrow4⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-3;-1;-4;0;-6;2\right\}\)

Loại những giá trị \(\sqrt{x}\in\left\{-3;-1;-4;-6;2\right\}\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\)

Vậy để P là số nguyên \(\Leftrightarrow x=0\)

Cho mình sửa 1 chút nhé :

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(N=\dfrac{xy\left(x^{\dfrac{1}{3}}+y^{\dfrac{1}{3}}\right)}{x^{\dfrac{1}{3}}+y^{\dfrac{1}{3}}}=xy\)

Bài làm:

Ta có: 

\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)

\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)

\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)