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a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b: \(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5\sqrt{6}}{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
\(a.\left(\dfrac{\sqrt{6}-\sqrt{3}}{5\sqrt{2}-5}+\dfrac{\sqrt{5}}{5}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}=\left[\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{5\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}}{5}\right]:\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{\sqrt{3}+\sqrt{5}}{5}.\dfrac{1}{\sqrt{5}+\sqrt{3}}=\dfrac{1}{5}\)
\(b.\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=\dfrac{-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=2\sqrt{3}-\dfrac{4}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}+1\right)-4}{\sqrt{3}+1}=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\left(1+\sqrt{3}\right)}{1+\sqrt{3}}=2\)
a. \(2\sqrt{16}+\sqrt{2}.\sqrt{0,02}-\dfrac{\sqrt{12,1}}{\sqrt{0,1}}=2.4+\sqrt{0,04}-\sqrt{\dfrac{12,1}{0,1}}=8+0,2-11=-2,8\)b. \(5\sqrt{20}-4\sqrt{45}+\dfrac{15}{\sqrt{5}}=10\sqrt{5}-12\sqrt{5}+3\sqrt{5}=\sqrt{5}\)
c. \(\left(\dfrac{\sqrt{6}-\sqrt{3}}{5\sqrt{2}-5}+\dfrac{\sqrt{5}}{5}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{5\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}}{5}\right).\dfrac{\sqrt{5}-\sqrt{3}}{2}=\dfrac{\sqrt{3}+\sqrt{5}}{5}.\dfrac{\sqrt{5}-\sqrt{3}}{2}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5.2}=\dfrac{5-3}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)d. \(\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=\dfrac{-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=3\sqrt{3}-\sqrt{3}-\dfrac{4}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}+1\right).2\sqrt{3}-4}{\sqrt{3}+1}=\dfrac{6+2\sqrt{3}-4}{\sqrt{3}+1}=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=\dfrac{ 2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)
a: \(=\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\cdot\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}=9\sqrt{3}-11\sqrt{2}\)
b: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{9-3}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
d: \(=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=-\sqrt{2}\)
a,
\(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\sqrt{7}+12}{3}-\dfrac{20-5\sqrt{7}}{9}\)
\(=\dfrac{2\sqrt{7}-8}{2}+\dfrac{18\sqrt{7}+36}{9}-\dfrac{20-5\sqrt{7}}{9}\)
\(=\sqrt{7}-4+\dfrac{23\sqrt{7}+16}{9}\)
\(=\dfrac{9\sqrt{7}-36}{9}+\dfrac{23\sqrt{7}+16}{9}=\dfrac{32\sqrt{7}-20}{9}\)
a: \(=2\cdot4+0.2-11=8-11+0.2=-2.8\)
b: \(=10\sqrt{5}-12\sqrt{5}+3\sqrt{5}=\sqrt{5}\)
c: \(=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{5\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}}{5}\right)\cdot\dfrac{\sqrt{5}-\sqrt{3}}{2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{5}\cdot\dfrac{\sqrt{5}-\sqrt{3}}{2}=\dfrac{1}{2}\)
d: \(=-\sqrt{3}+2-2\sqrt{3}+3\sqrt{3}=2\)
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b:\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5}{6}\sqrt{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
c: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)
\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
a: \(=4\sqrt{5}-\left|2-\sqrt{5}\right|+\sqrt{\dfrac{16}{5}}\)
\(=4\sqrt{5}-\sqrt{5}+2+\dfrac{4}{5}\sqrt{5}\)
\(=\dfrac{19}{5}\sqrt{5}+2\)
b: \(=\dfrac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(6+\sqrt{3}\right)}{\sqrt{3}}-\dfrac{13\left(4-\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}\)
\(=-\sqrt{3}+6+\sqrt{3}-4+\sqrt{3}=2+\sqrt{3}\)
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