Cho biểu thức B = 2(x + 4)/x - 3 căn bậc hai của x - 4 + căn bậc hai của x

* Hướng dẫn giải

$\begin{array}{l}a)B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\\ =\frac{2x+8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}+\frac{\sqrt{x}.\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\end{array}$

$\begin{array}{l}=\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\\ =\frac{3\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{3\sqrt{x}}{\sqrt{x}+1}\end{array}$

$b)B=\frac{3\sqrt{x}}{\sqrt{x}+1}=\frac{3\sqrt{x}+3-3}{\sqrt{x}+1}=3-\frac{3}{\sqrt{x}+1}$

Để $B\in \mathbb{Z}\Rightarrow 3⋮\left(\sqrt{x}+1\right)\Rightarrow \sqrt{x}+1\in U\left(3\right)=\left\{1;3\right\}$ do $\sqrt{x}+1>0$

$\Rightarrow \sqrt{x}\in \left\{0;2\right\}\Rightarrow x\in \left\{0;4\right\}$