Với x > = 0 ; x khác 4 

\(A=\dfrac{9+2x+\sqrt{x}-10-x+1}{x-\sqrt{x}-2}=\dfrac{x+\sqrt{x}}{x-\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{3}+2\)

\(=\sqrt{3}+2+\sqrt{2}-\sqrt{3}+2=4+\sqrt{2}\)

 tham khảo

ĐK: \(x>0;x\ne1\)

a) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\\ =x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)

b) \(P=x-\sqrt{x}+1=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu = xảy ra \(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\) (TM)

Vậy \(P_{min}=\dfrac{3}{4}\) đạt được khi \(x=\dfrac{1}{4}\).

c) \(Q=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\) \(\Rightarrow Q>0\)

Ta có \(Q-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}< 0\)

\(\Rightarrow Q< 2\)

a) \(ĐKXĐ:x\ge0;x\ne1\)

 \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left[\left(\sqrt{x}\right)^3-1\right]}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

b) Với \(x\ge0;x\ne1\), ta có:

\(P=x-\sqrt{x}+1=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

- Dấu "=" xảy ra khi \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{4}\)

- Vậy \(MinP=\dfrac{3}{4}\)

c) \(Q=\dfrac{2\sqrt{x}}{P}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)

\(Q< 2\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}< 2\)

\(\Leftrightarrow2\sqrt{x}< 2\left(x-\sqrt{x}+1\right)\)

\(\Leftrightarrow2\sqrt{x}< 2x-2\sqrt{x}+2\)

\(\Leftrightarrow2x-4\sqrt{x}+2>0\)

\(\Leftrightarrow2\left(\sqrt{x}-1\right)^2>0\) (đúng do \(x\ne1\))

- Vậy ta có đpcm.

- Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\ge0\\\sqrt{y-2}=b\ge0\\\sqrt{z-2}=c\ge0\end{matrix}\right.\)

- Đẳng thức đã cho tương đương:

\(a^2+b^2+c^2+3=2a+2b+2c\)

\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-2}=1\\\sqrt{z-2}=1\end{matrix}\right.\)

\(\Rightarrow x=y=z=3\)

\(Q=\sqrt{\left(x-y+1\right)^{2012}}+\sqrt{\left(x-3\right)^{2014}}+\sqrt{\left(y-4\right)^{2016}}\)

\(=\sqrt{1^{2012}}+\sqrt{\left(3-3\right)^{2014}}+\sqrt{\left(3-4\right)^{2016}}\)

\(=\sqrt{1}+\sqrt{0}+\sqrt{1}=2\)