ĐK: x>0,\(x\ne1\)
a) \(Q=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có Q=\(x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow Q\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy GTNN của Q là \(\dfrac{3}{4}\) và xảy ra khi \(x=\dfrac{1}{4}\)
c)
Ta có \(\dfrac{3Q}{\sqrt{x}}=\dfrac{3\left(x-\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{3x-3\sqrt{x}+3}{\sqrt{x}}=3\sqrt{x}-3+\dfrac{3}{\sqrt{x}}\)Vậy để \(\dfrac{3Q}{\sqrt{x}}\) nguyên thì \(\left\{{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\inƯ\left(3\right)\in\left(1;3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
Vậy x=9 thì \(\dfrac{3Q}{\sqrt{x}}\) nhận giá trị nguyên