Bài 2
Với a > 0 ; a khác 1 ; 2
\(P=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\right):\dfrac{a+2}{a-2}=2:\dfrac{a+2}{a-2}=\dfrac{2a-4}{a+2}\)
b, Để P nguyên \(\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
| a+2 | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 |
| a | -1 | -3 | 0(loại) | -4 | 2 | -6 | 6 | -10 |
Bài 3
a, Với x >= 0 ; x khác 9
\(A=\left(\dfrac{\sqrt{x}+3-\sqrt{x}+3}{x-9}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{x-9}:\dfrac{3}{\sqrt{x}-3}=\dfrac{2}{\sqrt{x}+3}\)
b, Ta có \(A>\dfrac{1}{3}\Rightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\Leftrightarrow\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
Do \(3\left(\sqrt{x}+3\right)>0\Rightarrow3-\sqrt{x}>0\Leftrightarrow x< 9\)
Kết hợp đk vậy 0 =< x < 9
