\(a.\sqrt{\frac{2-\sqrt{3}}{2}}+\frac{1-\sqrt{3}}{2}\)

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

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

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

\(=\frac{\sqrt{3}-1+1-\sqrt{3}}{2}\) ( Vì \(\sqrt{3}-1>0\))

\(=0\)

b) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)

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

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

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

\(=\frac{3+\sqrt{3}}{3}=\frac{\sqrt{3}+1}{\sqrt{3}}\)

c) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)

\(=\frac{2\left(2-\sqrt{3}\right)}{1}+\frac{13\left(1+\sqrt{3}\right)}{13}+2\sqrt{3}\)

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

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

ban mai thanh xuân ơi cầu c sai

cau c phai la \(8-\sqrt{3}\)

\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)

\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)

\(=\sqrt{\sqrt{5}^2+2\cdot\sqrt{5}\left(\sqrt{2}+1\right)+\left(\sqrt{2}+1\right)^2}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}\)

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

\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)

lam gjup vs mn oi

1/ ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

\(A=\left[\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)

\(=\left[\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{6}{\sqrt{x}+2}\right)\)

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

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

2/ Để \(A>2\Rightarrow\frac{-1}{3\left(\sqrt{x}-2\right)}>2\)\(\Rightarrow6\sqrt{x}-12+1>0\Rightarrow6\sqrt{x}-11>0\Rightarrow\sqrt{x}>\frac{11}{6}\)

                             \(\Rightarrow x>\frac{121}{36}\)

sai r bn oi

Ta có: \(\sqrt{2+\sqrt{3}}=\frac{1}{\sqrt{2}}.\sqrt{4+2\sqrt{3}}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}\)

=> \(A=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}}=\frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}}=\frac{\sqrt{3}+1}{2}\)

\(12\sqrt{\frac{4}{3}}-\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{4-6\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{3}-2}\)

\(=12.\frac{2}{\sqrt{3}}-\frac{\left(3+\sqrt{2}\right)\left(8-2\sqrt{2}\right)}{9-2}-\frac{\sqrt{2}\left(4-6\sqrt{2}\right)}{2}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{3-4}\)

\(=8\sqrt{3}-\left(4+2\sqrt{2}\right)-\left(2\sqrt{2}-6\right)+\left(-3-2\sqrt{3}\right)\)

\(=8\sqrt{3}-4-2\sqrt{2}-2\sqrt{2}+6-3-2\sqrt{3}\)

\(=6\sqrt{3}-4\sqrt{2}-1\)

giải được chết liền

what the hell

??????????????