Cho biểu thức P = 3 căn bậc hai x +2/ căn bậc hai x +1/ 2 căn bậc hai x -3/ 3 - căn bậc hai x - 3(3 căn bậc hai x-5)/x-2 căn bậc hai x -3

Điều kiện: $x\ge 0,x\ne 9$.

a) Với $x\ge 0,x\ne 9$ ta có:

$$\begin{gathered} P=\frac{3\sqrt{x}+2}{\sqrt{x}+1}-\frac{2\sqrt{x}-3}{3-\sqrt{x}}-\frac{3\left(3\sqrt{x}-5\right)}{(\sqrt{x}+1)(\sqrt{x}-3)} \\ =\frac{\left(3\sqrt{x}+2\right)\left(3-\sqrt{x}\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)-3\left(3\sqrt{x}-5\right)}{(\sqrt{x}+1)(\sqrt{x}-3)} \\ =\frac{3x-9\sqrt{x}+2\sqrt{x}-6+2x+2\sqrt{x}-3\sqrt{x}-3-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)} \\ =\frac{5x-17\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{5x-15\sqrt{x}-2\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}-3\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{5\sqrt{x}-2}{\sqrt{x}+1} \end{gathered}$$