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Ta có:
\(B=\frac{\sqrt{2+\sqrt{3}}}{2}\div\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(B=\frac{\sqrt{4+2\sqrt{3}}}{2}\div\left(\frac{\sqrt{4+2\sqrt{3}}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}\right)\)
\(B=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}\div\left(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{3}}\right)\)
\(B=\frac{\sqrt{3}+1}{2}\div\left(\frac{\sqrt{3}+1}{2}-\frac{2\sqrt{3}}{2}+\frac{\left(\sqrt{3}+1\right)\sqrt{3}}{6}\right)\)
\(B=\frac{\sqrt{3}+1}{2}\div\left[\frac{3\left(\sqrt{3}+1\right)-6\sqrt{3}+3+\sqrt{3}}{6}\right]\)
\(B=\frac{\sqrt{3}+1}{2}\div\frac{6-2\sqrt{3}}{6}\)
\(B=\frac{\sqrt{3}+1}{2}.\frac{6}{6-2\sqrt{3}}\)
\(B=\frac{3+2\sqrt{3}}{2}\)
a) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{\left(3-\sqrt{2}\right)+\left(3+\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-\left(\sqrt{2}\right)^2}=\frac{6}{7}\)
b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}=\frac{2\left(2\sqrt{3}+3\sqrt{3}\right)-3\left(3\sqrt{2}-3\sqrt{3}\right)}{\left(3\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{3}\right)}=\frac{19\sqrt{3}-9\sqrt{2}}{-45+15\sqrt{6}}=-\frac{13\sqrt{3}+10\sqrt{2}}{15}\)c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{2}=\frac{16}{2}=8\)d) \(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}=\frac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=-\frac{18\sqrt{3}}{19}\)
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) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)
\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)
\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)
\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)
\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)
\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)
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) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)+8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
= \(\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)
= \(\frac{2\left(5-\sqrt{5}+\sqrt{10}-\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)
= -2
b); c); d) làm tương tự
\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\) + \(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
<=> \(\frac{\left(\sqrt{2-\sqrt{3}}\right)^2+\left(\sqrt{2+\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2-\sqrt{3}}\right)}\)
<=>\(\frac{2-\sqrt{3}+2-\sqrt{3}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
<=>\(\frac{4}{\sqrt{4-3}}\)
<=> 4
mình năm nay lên lớp 9 nên có chỗ nào sai xót thì bạn sửa lại nha k mình nhé ^^