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\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne\pm2\end{cases}}\)
\(P=\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{2}{\sqrt{x}+2}-\frac{4\sqrt{x}}{x-4}\)
\(\Leftrightarrow P=\frac{x+2\sqrt{x}-2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
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 :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a) Ta có:
\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-4}{\sqrt{x}-2\sqrt{x}}\)
\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}+\frac{\sqrt{x}-4}{\sqrt{x}}\)
\(A=\frac{\left(\sqrt{x}-3\right)\sqrt{x}+\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\sqrt{x}}\)
\(A=\frac{x-3\sqrt{x}+x-6\sqrt{x}+8}{\left(\sqrt{x}-2\right)\sqrt{x}}\)
\(A=\frac{2x-9\sqrt{x}+8}{\left(\sqrt{x}-2\right)\sqrt{x}}\)
Bài làm:
Ta có:
\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)
\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)