Bài 1 :
1, Thay x = 36 vào A ta được : \(A=\dfrac{4\sqrt{36}+1}{\sqrt{36}+3}=\dfrac{25}{9}\)
2, Với \(x\ge0;x\ne9\)
\(B=\dfrac{x-\sqrt{x}+12}{x-9}-\dfrac{3}{\sqrt{x}-3}=\dfrac{x-\sqrt{x}+12-3\left(\sqrt{x}+3\right)}{x-9}\)
\(=\dfrac{x-4\sqrt{x}+3}{x-9}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
3, \(B< \dfrac{1}{5}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-\dfrac{1}{5}< 0\Leftrightarrow\dfrac{5\sqrt{x}-5-\sqrt{x}-3}{5\left(\sqrt{x}+3\right)}< 0\)
\(\Rightarrow4\sqrt{x}-8< 0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)
Kết hợp với đk vậy 0 =< x < 4
4, \(M=A:B=\dfrac{4\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{4\left(\sqrt{x}-1\right)+5}{\sqrt{x}-1}=4+\dfrac{5}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
| \(\sqrt{x}-1\) | 1 | -1 | 5 | -5 |
| x | 4 | 0 | 36 | loại |
