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

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

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

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

a) \(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}=0\)

b) \(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}=\frac{\left(5-2\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=\frac{\left(5\sqrt{2}+5\sqrt{3}-4\sqrt{3}-6\sqrt{2}\right)}{-1}\)

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

a,\(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}\)

\(=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}\)

\(=0\)

b,\(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}\)

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

\(=\frac{\sqrt{3}-\sqrt{2}}{-1}\)

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

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

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

=>A^2=24-8căn 3+2căn 3

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

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

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

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

\(M=10-3\sqrt{5}+\sqrt{45}\)

\(M=10-3\sqrt{5}+3\sqrt{5}\)

\(M=10\)

M = \(\sqrt{5.20}-\sqrt{5.3}+\sqrt{45}=\sqrt{100}-\sqrt{45}+\sqrt{45}=\sqrt{100}=10\)

Lời giải:
a.

$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$

b.

$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$

a: \(2\sqrt{45}+\sqrt{5}-3\sqrt{80}\)

\(=6\sqrt{5}+\sqrt{5}-12\sqrt{5}\)

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

b: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)

\(=2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}\)

\(=-8\sqrt{3}+1\)

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

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

*\(A=2\sqrt{80\sqrt{7}}-2\sqrt{45\sqrt{7}}-5\sqrt{20\sqrt{7}}\)

\(A=16\sqrt{5\sqrt{7}}-6\sqrt{5\sqrt{7}}-10\sqrt{5\sqrt{7}}\)

\(A=\left(16-6-10\right)\sqrt{5\sqrt{7}}=0\)

* \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right).\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\)

\(B^3=10-9B\)

\(\Rightarrow B^3+9B-10=0\)

\(\Rightarrow B^3-B^2+B^2-B+10B-10=0\)

\(\Rightarrow B^2\left(B-1\right)+B\left(B-1\right)+10\left(B-1\right)=0\)

\(\Rightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Rightarrow B=1\)