a) B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}+2}\) ( đk: x \(\ge\) 0; x\(e\)4)
<=> B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
<=> B = \(\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
<=> B = \(\dfrac{x+7\sqrt{x}+6}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
=> B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)
b) Để B > -1 => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}>-1\)
=> \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}+1>0\)
<=> \(\dfrac{-8}{\sqrt{x}-2}\) > 0 => \(\sqrt{x}-2< 0\) => \(x< 4\)
Đối chiếu với điều kiện ta được: \(0\le x< 4\)
c) Để B \(\in\) Z => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) \(\in\) Z
Mà \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) = \(\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}=-1-\dfrac{8}{\sqrt{x}-2}\)
=> \(\dfrac{8}{\sqrt{x}-2}\in Z\) => 8 \(⋮\) \(\sqrt{x}-2\)
=> \(\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
=> \(\sqrt{x}\in\left\{3;1;4;0;6;-4;10;-6\right\}\)
Mà \(x\ge0\) => \(x\in\left\{9;0;1;16;36;100\right\}\)
Vậy ====-...
