a, B= \(\frac{2\sqrt{x}+1}{x-7\sqrt{x}+12}-\frac{\sqrt{x}+3}{\sqrt{x}-4}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
<=> \(B=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-4}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
Để B có nghĩa
<=> \(\left\{{}\begin{matrix}\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)\ne0\\x\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\sqrt{x}\ne4\\\sqrt{x}\ne3\\x\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ne16\\x\ne9\\x\ge0\end{matrix}\right.\)
<=> \(x\ge0,x\ne16,x\ne9\)
Vậy để B có nghĩa <=> \(x\ge0,x\ne16,x\ne9\)
b, Có B=\(\frac{2\sqrt{x}+1}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-4}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)( đk: x\(\ge0\), \(x\ne16,x\ne9\))
<=> \(B=\frac{2\sqrt{x}+1-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}+1-x+9+2x-8\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)=\(\frac{x-5\sqrt{x}+6}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}=\frac{x-2\sqrt{x}-3\sqrt{x}+6}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}-2}{\sqrt{x}-4}\)