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:

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

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

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

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

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

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

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

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

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

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

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

(x-4,5)⁴(x-5,5)⁴ = 0

\(\Rightarrow\left[{}\begin{matrix}x-4,5=0\\x-5,5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4,5\\x=5,5\end{matrix}\right.\)

Điều kiện: \(x\ge1\)

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

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

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

$\iff \sqrt{x-1}+1=2$

$\iff$$\sqrt{x-1}=1$

$\iff {\left(\sqrt{x-1}\right)}^2=1^2$

$\iff x-1=1$

$\iff x=2$

\(\sqrt{x^2}=7\Rightarrow\left|x\right|=7\Rightarrow x=\pm7\)

x=7