\(A^2=12-\sqrt{80-32\sqrt{3}}+12+\sqrt{80-32\sqrt{3}}-2\sqrt{144-80+32\sqrt{3}}\)

=>\(A^2=24-2\sqrt{48+32\sqrt{3}}\)

=>A^2=24-8căn 3+2căn 3

=>\(A=\sqrt{24-8\sqrt{3+2\sqrt{3}}}\)

\(x=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}=\dfrac{1}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}\)

Đặt \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)\(\Leftrightarrow A^3=18+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow A^3=18+3A\sqrt[3]{1}\\ \Leftrightarrow A^3-3A-18=0\\ \Leftrightarrow A=3\\ \Leftrightarrow X=\dfrac{1}{3}\\ \Leftrightarrow Q=\left[3\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^2-1\right]^{2021}=\left(\dfrac{1}{9}-\dfrac{1}{9}-1\right)^{2021}=\left(-1\right)^{2021}=-1\)

!@#$%^&*()_+\ [];'{}

đầu hàng tại chỗ !

hiiiii

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

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

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

thay vao Q ta co

Q= \(-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{2012}+\sqrt{2013}=-\sqrt{2}+\sqrt{2013}\)

T = \(\dfrac{\sqrt{5}\left(\sqrt{16}-\sqrt{9}\right)}{4-5}-5\sqrt{5}+\dfrac{1}{\sqrt{5}-2}+2\sqrt{5}\)

   = \(-\sqrt{5}-5\sqrt{5}+2\sqrt{5}+\dfrac{1}{\sqrt{5}-2}\)

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

   = \(\dfrac{-4\sqrt{5}\left(\sqrt{5}-2\right)+1}{\sqrt{5}-2}\)

   = \(\dfrac{-20+8\sqrt{5}+1}{\sqrt{5}-2}\)

   = \(\dfrac{-19+8\sqrt{5}}{\sqrt{5}-2}\)

   = \(\dfrac{19-8\sqrt{5}}{2-\sqrt{5}}\)

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

Ta có: \(T=\dfrac{\sqrt{80}-\sqrt{45}}{4-\sqrt{25}}-\sqrt{125}+\dfrac{1}{\sqrt{5}-2}+\dfrac{2\sqrt{55}}{\sqrt{11}}\)

\(=\dfrac{4\sqrt{5}-3\sqrt{5}}{-1}-5\sqrt{5}+\sqrt{5}+2+2\sqrt{5}\)

\(=3\sqrt{5}-4\sqrt{5}-5\sqrt{5}+\sqrt{5}+2+2\sqrt{5}\)

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

1: Khi x=36 thì \(A=\dfrac{6}{2\cdot6-4}=\dfrac{6}{12-4}=\dfrac{6}{8}=\dfrac{3}{4}\)

2: 

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >4\end{matrix}\right.\)

\(C=B:A\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{3\sqrt{x}-x}{x-4}\right):\dfrac{\sqrt{x}}{2\sqrt{x}-4}\)

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

\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}-x}{\sqrt{x}+2}\cdot\dfrac{2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+2}\)

3: \(C\cdot\sqrt{x}< \dfrac{4}{3}\)

=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{3}< 0\)

=>\(\dfrac{2\sqrt{x}\cdot3-4\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)

=>\(6\sqrt{x}-4\sqrt{x}-8< 0\)

=>\(2\sqrt{x}-8< 0\)

=>\(\sqrt{x}< 4\)

=>\(0< =x< 16\)

Kết hợp ĐKXĐ của C, ta được: \(\left\{{}\begin{matrix}0< x< 16\\x< >4\end{matrix}\right.\)

\(A=5\sqrt{\dfrac{1}{5}}+\dfrac{5}{2}\cdot\sqrt{20}-\sqrt{80}\)

\(=\dfrac{5}{\sqrt{5}}+\dfrac{5}{2}\cdot2\sqrt{5}-4\sqrt{5}\)

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

a: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

Khi x=25 thì \(A=\dfrac{5+2}{5+3}=\dfrac{7}{8}\)

b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}+2}+\dfrac{x+4}{4-x}\)

\(=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-x-4}{x-4}\)

\(=\dfrac{5\sqrt{x}-10}{x-4}=\dfrac{5}{\sqrt{x}+2}\)

c: \(A\cdot B=\dfrac{5}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{5}{\sqrt{x}+3}\)

Để A*B>1 thì \(\dfrac{5}{\sqrt{x}+3}-1>0\)

=>\(\dfrac{5-\sqrt{x}-3}{\sqrt{x}+3}>0\)

=>\(2-\sqrt{x}>0\)

=>căn x<2

=>0<=x<4