Cho biểu thức B = (căn bậc hai x +1/ căn bậc hai x-2 -2 căn bậc hai x/ căn bậc hai x +2 + 5 căn bậc hai x+2/4-x)

Điều kiện: $x>0,x\ne 4,x\ne 9$.

a) Với $x>0,x\ne 4,x\ne 9$ ta có:

$$\begin{gathered} B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{5\sqrt{x}+2}{(2+\sqrt{x})(2-\sqrt{x})}\right):\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{{\left(\sqrt{x}+2\right)}^2} \\ =\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{{\left(\sqrt{x}+2\right)}^2}{\sqrt{x}\left(3-\sqrt{x}\right)} \\ =\frac{x+2\sqrt{x}+\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{{\left(\sqrt{x}+2\right)}^2}{\sqrt{x}\left(3-\sqrt{x}\right)} \\ =\frac{-x+2\sqrt{x}}{\sqrt{x}-2}.\frac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}=\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-2}.\frac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}=\frac{\sqrt{x}+2}{\sqrt{x}-3} \end{gathered}$$

Vậy $B=\frac{\sqrt{x}+2}{\sqrt{x}-3}$ với $x>0,x\ne 4,x\ne 9$.