a: \(E=1+1=2\)

b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)

\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)

d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)

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a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)

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

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

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

\(=2\sqrt{3}\)

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

\(=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{2}\)

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

\(=0\)

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

\(=\left[2-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\right]\cdot\left[2-\dfrac{\sqrt{5}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\right]\)

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

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

\(=9-4\sqrt{5}\)

d) \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

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

\(=6-121\)

\(=-115\)

\(2,\\ a,PT\Leftrightarrow\sqrt{\left(5x-1\right)^2}=\sqrt{4\left(x+1\right)^2}\\ \Leftrightarrow\left|5x-1\right|=2\left|x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=2\left(x+1\right)\\1-5x=2\left(x+1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\7x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{7}\end{matrix}\right.\)

\(b,ĐK:x^2-3\ge0\\ PT\Leftrightarrow\sqrt{x^2-3}=x-1\\ \Leftrightarrow x^2-3=x^2-2x+1\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ c,ĐK:x\le\dfrac{7}{2}\\ PT\Leftrightarrow7-2x=x^2+7\\ \Leftrightarrow x^2+2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge3\\ PT\Leftrightarrow3\sqrt{x-3}+\dfrac{1}{2}\cdot2\sqrt{x-3}-9\cdot\dfrac{1}{3}\sqrt{x-3}=2\\ \Leftrightarrow\sqrt{x-3}=2\\ \Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)

Bài 1: 

d: Ta có: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{20}{5+\sqrt{5}}-\sqrt{20}\)

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

=-3

1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

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

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

2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)

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

\(=4\sqrt{2}\)

\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)

\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)

\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)

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

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

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

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

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

\(------\)

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

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

\(=\dfrac{4\sqrt{5}-4}{5-1}+\dfrac{5\sqrt{5}-10}{5-4}+\dfrac{5\sqrt{5}-15}{5-9}\)

\(=5\sqrt{5}-10+\left(\dfrac{4\sqrt{5}-4}{4}+\dfrac{5\sqrt{5}-15}{-4}\right)\)

\(=\dfrac{4\cdot\left(5\sqrt{5}-10\right)}{4}+\left(\dfrac{4\sqrt{5}-4}{4}-\dfrac{5\sqrt{5}-15}{4}\right)\)

\(=\dfrac{20\sqrt{5}-40}{4}+\dfrac{-\sqrt{5}+11}{4}\)

\(=\dfrac{19\sqrt{5}-29}{4}\)

#Ayumu

5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)

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

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