\(\frac{1}{\sqrt{1}+\sqrt{5}}\)+\(\frac{1}{\sqrt{9}+\sqrt{13}}\)+\(\frac{1}{\sqrt{13}+\sqrt{17}}+\frac{1}{\sqrt{17}+\sqrt{21}}\)+\(\frac{1}{\sqrt{23}+\sqrt{25}}\)
K
Khách
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Những câu hỏi liên quan
8 tháng 9 2019
Nếu đề đúng:
Sử dụng liên hợp để trục căn thức ở mẫu:
\(\frac{1}{\sqrt{1}+\sqrt{5}}=\frac{\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{\sqrt{5}-1}{5-1}=\frac{\sqrt{5}-1}{4}\)
Tương tự như vậy ta sẽ có:
\(N=\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\sqrt{13}-\sqrt{9}}{\left(\sqrt{13}-\sqrt{9}\right)\left(\sqrt{13}+\sqrt{9}\right)}+\frac{\sqrt{17}-\sqrt{13}}{\left(\sqrt{17}-\sqrt{13}\right)\left(\sqrt{17}+\sqrt{13}\right)}\)
\(+\frac{\sqrt{21}-\sqrt{17}}{\left(\sqrt{21}-\sqrt{17}\right)\left(\sqrt{21}+\sqrt{17}\right)}+\frac{\sqrt{25}-\sqrt{23}}{\left(\sqrt{25}-\sqrt{23}\right)\left(\sqrt{25}+\sqrt{23}\right)}\)
\(=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+\frac{\sqrt{17}-\sqrt{13}}{4}+\frac{\sqrt{21}-\sqrt{17}}{4}+\frac{\sqrt{25}-\sqrt{23}}{4}\)
\(=\frac{\sqrt{5}-1+\sqrt{13}-\sqrt{9}+\sqrt{17}-\sqrt{13}+\sqrt{21}-\sqrt{17}+\sqrt{25}-\sqrt{23}}{4}\)
\(=\frac{\sqrt{5}-1-\sqrt{9}+\sqrt{21}+\sqrt{25}-\sqrt{23}}{4}=\frac{\sqrt{5}-1-3+\sqrt{21}+5-\sqrt{23}}{4}=\frac{1+\sqrt{5}+\sqrt{21}-\sqrt{23}}{4}\)
10 tháng 10 2017
Ta có:
\(\frac{1}{n\sqrt{n+4}+\left(n+4\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+4\right)}\left(\sqrt{n}+\sqrt{n+4}\right)}\)
\(=\frac{\sqrt{n+4}-\sqrt{n}}{4\sqrt{n\left(n+4\right)}}=\frac{1}{4}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+4}}\right)\)
Áp dụng vào bài toán ta được
\(\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)
\(=\frac{1}{4}.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2013}}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{\sqrt{2013}}\right)\)
8 tháng 8 2017
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
27 tháng 5 2018
\(A=\frac{1}{\sqrt{1}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2014}+\sqrt{2018}}\)
\(\Rightarrow A=\sqrt{5}-\sqrt{1}+\sqrt{9}-\sqrt{5}+...+\sqrt{2018}-\sqrt{2014}\)
\(\Rightarrow A=-\sqrt{1}+\sqrt{2018}\)
cho mk nha
Ai trên 11 điểm cho mình nha câu dưới 3 mk lại
BT
28 tháng 5 2018
\(Q=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
=> \(Q=\frac{1-\sqrt{5}}{-4}+\frac{\sqrt{5}-\sqrt{9}}{-4}+\frac{\sqrt{9}-\sqrt{13}}{-4}+...+\frac{\sqrt{2001}-\sqrt{2005}}{-4}\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}\right)\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{2005}\right)\)
=> \(Q=\frac{\sqrt{2005}-1}{4}\)
2 tháng 10 2018
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{2}}\right)\left(4-\sqrt{10+2\sqrt{2}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5.1}+1}=8+2\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(B=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=-\frac{1}{4}\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+....+\sqrt{2001}-\sqrt{2005}\right)\)
\(=-\frac{1}{4}\left(1-\sqrt{2005}\right)\)
\(=10,94430659\)
\(\text{Lm hơi vắn tắt thông cảm nha!!}\)
TT
2 tháng 4 2017
Ôi, trang wed không tự nhận diện được công thức latex. Mình đăng lại bài giải:
a) Ta có
\(4T=\frac{4}{1+\sqrt{5}}+\frac{4}{\sqrt{5}+\sqrt{9}}+...+\frac{4}{\sqrt{2013}+\sqrt{2017}}\)
\(=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{\sqrt{5}+1}+...+\frac{\left(\sqrt{2017}+\sqrt{2013}\right)\left(\sqrt{2017}-\sqrt{2013}\right)}{\sqrt{2017}+\sqrt{2013}}\)
\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+\sqrt{13}-\sqrt{9}+...+\sqrt{2017}-\sqrt{2013}\)
\(=\sqrt{2017}-1\)
\(\Rightarrow T=\frac{\sqrt{2017}-1}{4}\)
b) Ta có
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{2-1}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Tương tự ta có
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
......................
\(\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
Suy ra
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
TT
1 tháng 4 2017
a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5 + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013} + \sqrt {2017} }}\\
= \frac{{(\sqrt 5 + 1)(\sqrt 5 - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017} + \sqrt {2013} )(\sqrt {2017} - \sqrt {2013} )}}{{\sqrt {2013} + \sqrt {2017} }}\\
= \sqrt 5 - 1 + \sqrt 9 - \sqrt 5 + ... + \sqrt {2017} - \sqrt {2013} \\
= 1 + \sqrt 5 - \sqrt 5 + \sqrt 9 - \sqrt 9 + ... + \sqrt {2013} - \sqrt {2013} + \sqrt {2017} \\
= 1 + \sqrt {2017} \\
\Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]