Với `x >= 0,x \ne 4,x \ne 9` có:
`B=[2\sqrt{x}-9]/[x-5\sqrt{x}+6]-[\sqrt{x}+3]/[\sqrt{x}-2]+[2\sqrt{x}+1]/[\sqrt{x}-3]`
`B=[2\sqrt{x}-9-(\sqrt{x}+3)(\sqrt{x}-3)+(2\sqrt{x}+1)(\sqrt{x}-2)]/[(\sqrt{x}-2)(\sqrt{x}-3)]`
`B=[2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2]/[(\sqrt{x}-2)(\sqrt{x}-3)]`
`B=[x-\sqrt{x}-2]/[(\sqrt{x}-2)(\sqrt{x}-3)]`
`B=[(\sqrt{x}+2)(\sqrt{x}+1)]/[(\sqrt{x}-2)(\sqrt{x}-3)]=[\sqrt{x}+1]/[\sqrt{x}-3]`
\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(B=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne4\end{matrix}\right.\)
\(B=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
