\(E=\left(\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}+\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}-1}}\right):\sqrt{\frac{1}{x-1}}\) \(ĐKXĐ:x>1\)

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

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

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

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

\(=2\sqrt{x}\)

Ta có:\(x=19-8\sqrt{3}=16-2.4\sqrt{3}+3=\left(4-\sqrt{3}\right)^2\)

\(\Rightarrow2\sqrt{x}=2.\sqrt{\left(4-\sqrt{3}\right)^2}=2.\left(4-\sqrt{3}\right)=8-2\sqrt{3}\)

\(2\sqrt{3}-5=\sqrt{12}-5< \sqrt{25}-4=5-4\)

\(C=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x-\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\) (tự tìm ĐKXĐ)

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

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

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

\(=x-\sqrt{x}+3\)

GTNN:\(x-\sqrt{x}+3=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

\(\Rightarrow Min\left(C\right)=\frac{11}{4}khi..\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

TL:

\(\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\) 

\(=\frac{8-3\sqrt{7}-8-3\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)

\(=\frac{-6\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)

Cho   \(A=\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)

CACH  1  : \(\Rightarrow A\sqrt{2}=\sqrt{16-6\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)

\(\Rightarrow A\sqrt{2}=\sqrt{9-2.3.\sqrt{7}+7}-\sqrt{9+2.3.\sqrt{7}+7}\)

\(\Rightarrow A\sqrt{2}=\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(3+\sqrt{7}\right)^2}\)

\(\Rightarrow A\sqrt{2}=|3-\sqrt{7}|-|3+\sqrt{7}|\)

\(\Rightarrow A\sqrt{2}=3-\sqrt{7}-3-\sqrt{7}=-2\sqrt{7}=-\sqrt{28}\)

\(\Rightarrow A=-\sqrt{14}\)

CACH   2  :   \(A^2=8-3\sqrt{7}+8+3\sqrt{7}-2.\sqrt{8^2-\left(3\sqrt{7}\right)^2}\)

\(\Rightarrow A^2=16-2\sqrt{64-63}=16-2=14\)

\(\Rightarrow A=\sqrt{14}\) hoặc  \(A=-\sqrt{14}\)

Mà  \(8+3\sqrt{7}>8-3\sqrt{7}\) \(\Rightarrow\sqrt{8+3\sqrt{7}}>\sqrt{8-3\sqrt{7}}\)

Vây  A  âm  \(\Rightarrow A=-\sqrt{14}\)

a,ĐKXĐ:\(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\)

  \(\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)

 \(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\) 

\(\Rightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=4-\sqrt{x}-\sqrt{y}\left(đk:x;y>0\right)\)

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

Do x,y là các số thực dương nên sử dụng BĐT AM-GM cho 2 số không âm ta có :

\(\frac{1}{\sqrt{x}}+\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}.\sqrt{x}}=2\)

\(\frac{1}{\sqrt{y}}+\sqrt{y}\ge2\sqrt{\frac{1}{\sqrt{y}}.\sqrt{y}}=2\)

Cộng theo vế các bất đẳng thức cùng chiều ta được :

\(\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{y}}+\sqrt{y}\ge2+2=4\)

Dấu = xảy ra khi và chỉ khi \(\hept{\begin{cases}\frac{1}{\sqrt{x}}=\sqrt{x}\Leftrightarrow x=1\\\frac{1}{\sqrt{y}}=\sqrt{y}\Leftrightarrow y=1\end{cases}\Leftrightarrow}x=y=1\)

Vậy nghiệm của phương trình trên là \(x=y=1\)

\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)

\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)

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

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

\(=9-2=7\)

\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)

\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)

\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)

\(=\sqrt{3-\sqrt{3}}\)

\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)

\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)

\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)

\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)

\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)

\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)

\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)

\(=2.6-18\sqrt{2}+16\sqrt{3}\)

\(=12-18\sqrt{2}+16\sqrt{3}\)

\(F=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2012}+\sqrt{2011}}\)

\(F=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{2012}-\sqrt{2011}}{\left(\sqrt{2012}+\sqrt{2011}\right)\left(\sqrt{2012}-\sqrt{2011}\right)}\)

\(F=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2012}-\sqrt{2011}}{2012-2011}\)

\(F=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2012}-\sqrt{2011}\)

\(F=\sqrt{2012}-\sqrt{1}\)

\(F=\sqrt{2012}-1\)