\(\sqrt{1.2015}\le\frac{2016}{2}\Rightarrow\frac{1}{\sqrt{1.2015}}\ge\frac{2}{2016}\)

=>S\(\ge\frac{2.1015}{2016}\)\(>\frac{2.2014}{2015}\)

Bài này dễ dùng Cô si cho mẫu là OK

Ta thấy: k thuộc N^* nên \(\sqrt{k+1}>\sqrt{k}\)

\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{2}{\left(2\sqrt{k+1}\right).\left(\sqrt{k+1}.\sqrt{k}\right)}< \frac{2}{\left(\sqrt{k+1}.\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)}\)

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

\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)(đpcm).

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

 \(=\frac{\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k}\sqrt{k+1}}.\frac{\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}}<\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}.2\)  

Đề đúng  sory nhé

ĐKXĐ: a > 0

a/ \(K=\left[\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{1}{\sqrt{a}-1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)

       \(=\left[\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{\sqrt{a}+3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)

        \(=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+3}\right]\)  \(=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)

b/ Ta có: \(\sqrt{a}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

     \(K=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}=\frac{\left(\sqrt{2}+2\right)\sqrt{2}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}+4\right)}=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}\left(\sqrt{2}+1\right)\left(2\sqrt{2}+1\right)}\)

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

c/ \(K< 0\Leftrightarrow\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}< 0\)\(\Rightarrow\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)< 0\)

       \(\Rightarrow\sqrt{a}-1< 0\) (vì \(\left(\sqrt{a}+1\right)^2>0\))    \(\Rightarrow\sqrt{a}< 1\Rightarrow a< 1\)

               Vậy \(0< a< 1\) thì K < 0

a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)

\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a-1}\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\left(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

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

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

\(=\frac{\sqrt{a}+1}{\sqrt{a}}\cdot\sqrt{a}-1\)

\(=\frac{a-1}{\sqrt{a}}\)

b) thay \(a=3+2\sqrt{2}\) vào bt K được:

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

c) để K>0 thì:

\(\frac{a-1}{\sqrt{a}}>0\)

\(\Rightarrow a-1>0\)

\(\Rightarrow a>1\)

a) + b) ĐKXĐ : \(x>0,x\ne1\)

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

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

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

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

c) Để K âm thì : \(\frac{x-1}{\sqrt{x}}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp với ĐKXĐ \(\Leftrightarrow x>11\)

ĐKXĐ: \(x\ge0;x\ne1\)

\(K=\left(\frac{x+3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}-1\right)\)

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

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

\(K=\frac{2\left(\sqrt{x}+2\right)-5}{\sqrt{x}+2}=2-\frac{5}{\sqrt{x}+2}\)

\(\Rightarrow5⋮\left(\sqrt{x}+2\right)\) mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=5\Rightarrow x=9\)

c/ \(\frac{5}{\sqrt{x}+2}>0\Rightarrow2-\frac{5}{\sqrt{x}+2}< 2\Rightarrow K< 2\)

d/ \(\sqrt{x}+2\ge2\Rightarrow2-\frac{5}{\sqrt{x}+2}\ge2-\frac{5}{2}=-\frac{1}{2}\)

\(\Rightarrow K_{min}=-\frac{1}{2}\) khi \(x=0\)

e/ \(K=\frac{2\sqrt{x}-1}{\sqrt{x}+2}\Leftrightarrow K\sqrt{x}+2K=2\sqrt{x}-1\)

\(\Leftrightarrow\left(K-2\right)\sqrt{x}=-2K-1\Rightarrow\sqrt{x}=\frac{2K+1}{2-K}\)

Mà \(\sqrt{x}\ge0\Rightarrow\frac{2K+1}{2-K}\ge0\Rightarrow-\frac{1}{2}\le K< 2\)

\(\Rightarrow K=\left\{0;1\right\}\)

- Với \(K=0\Rightarrow\frac{2\sqrt{x}-1}{\sqrt{x}+2}=0\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)

- Với \(K=1\Rightarrow\frac{2\sqrt{x}-1}{\sqrt{x}+2}=1\Rightarrow2\sqrt{x}-1=\sqrt{x}+2\Rightarrow x=9\)