Lời giải:
1.
\(Q=\left[\frac{x}{x-\sqrt{x}}+\frac{x+1}{\sqrt{x}}-\frac{1}{\sqrt{x}-1}\right].\frac{x+25}{x+\sqrt{x}+1}\)
\(=\frac{x+(\sqrt{x}-1)(x+1)-\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}.\frac{x+25}{x+\sqrt{x}+1}=\frac{x\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)}.\frac{x+25}{x+\sqrt{x}+1}\)
\(=\frac{(\sqrt{x}-1)(x+\sqrt{x}+1)(x+25)}{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{x+25}{\sqrt{x}}\)
2.
Áp dụng BĐT Cô-si: $x+25\geq 10\sqrt{x}$
$\Rightarrow Q=\frac{x+25}{\sqrt{x}}\geq 10$
Vậy $Q_{\min}=10$ khi $x=25$
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