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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: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
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,
\(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\sqrt{7}+12}{3}-\dfrac{20-5\sqrt{7}}{9}\)
\(=\dfrac{2\sqrt{7}-8}{2}+\dfrac{18\sqrt{7}+36}{9}-\dfrac{20-5\sqrt{7}}{9}\)
\(=\sqrt{7}-4+\dfrac{23\sqrt{7}+16}{9}\)
\(=\dfrac{9\sqrt{7}-36}{9}+\dfrac{23\sqrt{7}+16}{9}=\dfrac{32\sqrt{7}-20}{9}\)
2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
\(=3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
\(=\left(\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\right)\left(4-2\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{5}\right)\left(4-2\sqrt{3}\right)\)
\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)
a: \(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{2+\sqrt{3}}{2}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{2-\sqrt{3}}{2}}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\dfrac{\sqrt{3}+1}{2}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\dfrac{\sqrt{3}-1}{2}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}\cdot\dfrac{2}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2}\cdot\dfrac{2}{2-\sqrt{3}+1}\)
\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)
\(=\dfrac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}\)
\(=\dfrac{6}{6}=1\)
a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
a. \(\sqrt{18}-\dfrac{1}{2}\sqrt{48}-\sqrt{8}+\dfrac{4-5\sqrt{2}}{5-2\sqrt{2}}\)
\(=3\sqrt{2}-2\sqrt{3}-2\sqrt{2}+\dfrac{\left(4-5\sqrt{2}\right)\left(2+5\sqrt{2}\right)}{\left(5-2\sqrt{2}\right)\left(5+2\sqrt{2}\right)}\)
\(=3\sqrt{2}-2\sqrt{3}-2\sqrt{2}+\dfrac{-17\sqrt{2}}{17}\)
\(=3\sqrt{2}-2\sqrt{3}-2\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{3}\)
b. \(\dfrac{\sqrt{15}-2\sqrt{3}}{2-\sqrt{5}}+6\sqrt{\dfrac{1}{3}}+\dfrac{13}{\sqrt{3}-4}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{2-\sqrt{5}}+2\sqrt{3}+\dfrac{13\left(\sqrt{3}+4\right)}{\left(\sqrt{3}-4\right)\left(\sqrt{3}+4\right)}\)
\(=\dfrac{-\sqrt{3}\left(\sqrt{5}-2\right)}{2-\sqrt{5}}+2\sqrt{3}+\dfrac{13\left(\sqrt{3}+4\right)}{-13}\)
\(=-\sqrt{3}+2\sqrt{3}-\sqrt{3}-4\)
\(=-4\)
Xong r nhae
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\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)
\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)
1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)
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, Sửa đề:
\(A=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-2-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-2+\sqrt{3}}\)
\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}-\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{2\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{2\sqrt{6-3\sqrt{3}}}{3}\)