Giải thích các bước giải:
a.Ta có:
$P=\dfrac{x^2-\sqrt{x} }{x+\sqrt{x} +1}-\dfrac{2x+\sqrt{x} }{\sqrt{x} }+\dfrac{2(x-1)}{\sqrt{x} -1}$
$\to P=\dfrac{\sqrt{x}((\sqrt{x})^3-1) }{x+\sqrt{x} +1}-\dfrac{\sqrt{x}(2\sqrt{x}+1) }{\sqrt{x} }+\dfrac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x} -1}$
$\to P=\dfrac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x} +1) }{x+\sqrt{x} +1}-(2\sqrt{x}+1) +2(\sqrt{x}+1)$
$\to P=\sqrt{x}(\sqrt{x}-1)-(2\sqrt{x}+1) +2(\sqrt{x}+1)$
$\to P=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2$
$\to P=x-\sqrt{x}+1$
$\to P=(\sqrt{x}-\dfrac12)^2+\dfrac34\ge \dfrac34$
Dấu = xảy ra khi $\sqrt{x}=\dfrac12\to x=\dfrac14$
b.Ta có:
$Q=\dfrac{2\sqrt{x}}{P}$
$\to Q=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}$
Từ câu a $\to x-\sqrt{x}+1>0$
Vì $x>0\to Q>0$
Ta có:
$Q-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2$
$\to Q-2=\dfrac{2\sqrt{x}-2(x-\sqrt{x}+1)}{x-\sqrt{x}+1}$
$\to Q-2=\dfrac{-2(x-2\sqrt{x}+1)}{x-\sqrt{x}+1}$
$\to Q-2=\dfrac{-2(\sqrt{x}-1)^2}{x-\sqrt{x}+1}\le 0$
$\to Q\le 2$
$\to 0<Q\le 2$
Mà $Q\in Z\to Q\in\{1, 2\}$
Nếu $Q=2\to \dfrac{-2(\sqrt{x}-1)^2}{x-\sqrt{x}+1}=0\to \sqrt{x}=1\to x=1$
Nếu $Q=1\to \dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1$
$\to x-\sqrt{x}+1=2\sqrt{x}$
$\to x=\dfrac{7\pm3\sqrt5}{2}$
$\to x\in\{1, \dfrac{7\pm3\sqrt5}{2}\}$
