Cho \(B=\left( \frac{x}{\sqrt{x}+3}-\frac{x+1}{\sq...

D

- Hướng dẫn giải

Phương pháp giải:

+) Xác định mẫu thức chung của

Giải chi tiết:

\(\begin{align}  B=\left( \frac{x}{\sqrt{x}+3}-\frac{x+1}{\sqrt{x}-3}+\frac{6x+\sqrt{x}}{x-9} \right):\left( \frac{\sqrt{x}-3}{\sqrt{x}+3}-1 \right)=\left( \frac{x}{\sqrt{x}+3}-\frac{x+1}{\sqrt{x}-3}+\frac{6x+\sqrt{x}}{\left( \sqrt{x}+3 \right)\left( \sqrt{x}-3 \right)} \right):\left( \frac{\sqrt{x}-3}{\sqrt{x}+3}-1 \right) \\  \,\,\,\,\,=\frac{x\left( \sqrt{x}-3 \right)-\left( x+1 \right)\left( \sqrt{x}+3 \right)+6x+\sqrt{x}}{\left( \sqrt{x}+3 \right)\left( \sqrt{x}-3 \right)}:\frac{\sqrt{x}-3-\sqrt{x}-3}{\sqrt{x}+3}=\frac{x\sqrt{x}-3x-x\sqrt{x}-3x-\sqrt{x}-3+6x+\sqrt{x}}{\left( \sqrt{x}+3 \right)\left( \sqrt{x}-3 \right)}.\frac{\sqrt{x}+3}{-6} \\  \,\,\,\,\,=\frac{-3}{\left( \sqrt{x}+3 \right)\left( \sqrt{x}-3 \right)}.\frac{\sqrt{x}+3}{-6}=\frac{1}{2\left( \sqrt{x}-3 \right)} \\ \end{align}\)

\(x=12+6\sqrt{3}=9+2.3.\sqrt{3}+3={{\left( 3+\sqrt{3} \right)}^{2}}\Rightarrow \sqrt{x}=3+\sqrt{3}\)

Khi đó \(B=\frac{1}{2\left( \sqrt{x}-3 \right)}=\frac{1}{2\left( 3+\sqrt{3}-3 \right)}=\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\)