\(A=\frac{\left(2\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\sqrt{5}+2\right)\left(\sqrt{5}+1\right)-\left(10+2\sqrt{5}\right)\left(\sqrt{5}-1\right)}{5-1}-1\)

\(=\frac{10+2\sqrt{5}+2\sqrt{5}+2-10\sqrt{5}+10-10+2\sqrt{5}}{4}-1\)

\(=\frac{12-4\sqrt{5}}{4}-1\)

\(=\frac{4\left(3-\sqrt{5}\right)}{4}-1\)

\(=3-\sqrt{5}-1\)

\(=2-\sqrt{5}\) 

(còn biểu thức B hình như sai đề, bạn coi lại đề)

đề câu B nè : \(B=\sqrt{\left(1-\sqrt{2014}\right)^2}.\sqrt{2015+2\sqrt{2014}}\)

1: Rút gọn biểu thức

a) Ta có: \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=5\cdot\frac{1}{\sqrt{5}}+\frac{1}{3}\cdot3\sqrt{5}+\left|2-\sqrt{5}\right|\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-2\)(Vì \(2< \sqrt{5}\))

\(=3\sqrt{5}-2\)

b) Ta có: \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)

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

\(=\frac{30+10\sqrt{5}+30-10\sqrt{5}}{25-5}\)

\(=\frac{60}{20}=3\)

2:

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

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

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

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

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

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

\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)

b) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4;\frac{14\pm6\sqrt{5}}{4}\right\}\end{matrix}\right.\)

Để \(A>\frac{1}{6}\) thì \(A-\frac{1}{6}>0\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}-\frac{1}{6}>0\)

\(\Leftrightarrow\frac{2\sqrt{x}-4}{6\sqrt{x}}-\frac{\sqrt{x}}{6\sqrt{x}}>0\)

\(\Leftrightarrow\frac{\sqrt{x}-4}{6\sqrt{x}}>0\)

mà \(6\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-4>0\)

\(\Leftrightarrow\sqrt{x}>4\)

hay x>16

Kết hợp ĐKXĐ, ta được: x>16

Vậy: Để \(A>\frac{1}{6}\)thì x>16

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé