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b/ Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{99}-\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
Cả 2 câu là n tự nhiên khác 0 hết nhé
a/ Ta có: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Áp đụng vào bài toán được
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{1681}-\sqrt{1680}\)
\(=\sqrt{1681}-\sqrt{1}=41-1=40\)
a) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\frac{\sqrt{3}.\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}.\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\sqrt{3}+2+\sqrt{2}-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}\)
b) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)
\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)
\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)
\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)
\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)
\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
a) \(4\sqrt{\frac{2}{9}}+\sqrt{2}+\sqrt{\frac{1}{18}}\)
\(=\frac{8\sqrt{2}}{6}+\frac{6\sqrt{2}}{6}+\frac{\sqrt{2}}{6}\)
\(=\frac{15\sqrt{2}}{6}=\frac{5\sqrt{2}}{2}\)
b) \(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+1}{3-1}-\frac{\sqrt{3}-1}{3-1}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}=1\)