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Answer:

a. \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)   ĐK: \(x\ge0;x\ne1\)

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

\(=\frac{-2\sqrt{x}}{\sqrt{x}+1}.\frac{x-1}{2}\)

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

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

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

b. Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\1-\sqrt{x}>0\end{cases}}\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Do vậy \(\sqrt{x}\left(1-\sqrt{x}\right)>0\)

c. \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)

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

\(=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)

\(=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)

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Đặt \(\sqrt{x-1}=a;\sqrt{x+1}=b\) \(\left(a;b\ge0;x\ge1\right)\)

\(\sqrt{\left(x-1\right)\left(x+1\right)}=\sqrt{x+1}+\sqrt{x-1}-x+4\)

<=> ab = a + b - x + 4

<=> 2ab = 2(a + b) - 2x + 8

<=> 2ab = 2(a + b) - a² - b² + 8

<=> (a + b)² - 2(a + b) + 1 = 9

<=> (a + b - 1)² = 9

<=> \(\orbr{\begin{cases}a+b=4\\a+b=-2\end{cases}}\Leftrightarrow a+b=4\)

\(\Leftrightarrow\sqrt{x+1}+\sqrt{x-1}=4\)

\(\Leftrightarrow\sqrt{x+1}=4-\sqrt{x-1}\)

\(\Leftrightarrow\hept{\begin{cases}x+1=x-1-8\sqrt{x-1}+16\\1\le x\le17\end{cases}}\Leftrightarrow\hept{\begin{cases}4\sqrt{x-1}=7\\1\le x\le17\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}16\left(x-1\right)=49\\1\le x\le17\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{65}{16}\\1\le x\le17\end{cases}}\Leftrightarrow x=\frac{65}{16}\left(tm\right)\)

ĐK : \(x>2009;y>2010;z>2011\)

PT <=> \(\frac{-4\sqrt{x-2009}+4}{x-2009}+\frac{-4\sqrt{y-2010}+4}{y-2010}+\frac{-4\sqrt{z-2011}+4}{z-2011}=-3\)

<=> \(\frac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}+\frac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}+\frac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}=0\)

<=> \(\hept{\begin{cases}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)

Vậy phương trình có 1 nghiệm duy nhất (x;y;z) = (2013 ; 2014 ; 2015)