Ta có \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\right)=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\left[\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right]=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)=\sqrt{x}+4\)
Để \(\frac{A}{B}\ge\frac{x}{4}+5\) thì \(\sqrt{x}+4\ge\frac{x}{4}+5\Leftrightarrow\sqrt{x}\ge\frac{x}{4}+1\Leftrightarrow x-4\sqrt{x}+4\le0\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)
Mà \(\left(\sqrt{x}-2\right)^2\ge0\)
Suy ra \(\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(tm)
Vậy x=4 thì \(\frac{A}{B}\ge\frac{x}{4}+5\)
\(B=\frac{1}{\sqrt{x}-1}\) (tự rút gọn nha)
\(\frac{A}{B}\ge\frac{x}{4}+5\\ \sqrt{x}+4\ge\frac{x}{4}+5\\ \frac{x}{4}-\sqrt{x}+1\le0\\ x-4\sqrt{x}+4\le0\\ \left(\sqrt{x}-2\right)^2\le0\\ \Rightarrow\sqrt{x}-2=0\\ \Rightarrow x=4\)
Vậy để \(\frac{A}{B}\ge\frac{x}{4}+5\) thì x=4
