tính tổng
\(S=\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+...+\frac{1}{\sqrt{2019}+\sqrt{2017}}\)
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11 tháng 7 2019
\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)
\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)
\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)
\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)
\(S=\frac{\left(2019-1\right)}{2}=1009\)
11 tháng 7 2019
\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)
\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)
\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)
\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)
24 tháng 6 2018
......................?
mik ko biết
mong bn thông cảm
nha ................
6 tháng 8 2019
\(S=\frac{1}{3\sqrt{1}+3\sqrt{3}}+\frac{1}{3\sqrt{3}+3\sqrt{5}}+...+\frac{1}{3\sqrt{2017}+3\sqrt{2019}}\)
\(S=\frac{1}{3}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{2017}+\sqrt{2019}}\right)\)
\(S=\frac{1}{3}\left[\frac{\sqrt{3}-\sqrt{1}}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+...+\frac{\sqrt{2019}-\sqrt{2017}}{2019-2017}\right]\)
\(S=\frac{1}{3}\cdot\frac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2017}}{2}\)
\(S=\frac{\sqrt{2019}-\sqrt{1}}{6}\)
Y
6 tháng 8 2019
\(2S=\frac{1}{3}\left(\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}+...+\frac{2}{\sqrt{2017}+\sqrt{2019}}\right)\)
\(2S=\frac{1}{3}\left(\frac{3-1}{\sqrt{1}+\sqrt{3}}+\frac{5-3}{\sqrt{3}+\sqrt{5}}+...+\frac{2019-2017}{\sqrt{2017}+\sqrt{2019}}\right)\)
\(2S=\frac{1}{3}\left(\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{3}+\sqrt{5}}+...+\frac{\left(\sqrt{2019}-\sqrt{2017}\right)\left(\sqrt{2019}+\sqrt{2017}\right)}{\sqrt{2019}+\sqrt{2017}}\right)\)
\(2S=\frac{1}{3}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2017}\right)\)
\(S=\frac{\sqrt{2019}-1}{6}\)
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)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5 + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013} + \sqrt {2017} }}\\
= \frac{{(\sqrt 5 + 1)(\sqrt 5 - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017} + \sqrt {2013} )(\sqrt {2017} - \sqrt {2013} )}}{{\sqrt {2013} + \sqrt {2017} }}\\
= \sqrt 5 - 1 + \sqrt 9 - \sqrt 5 + ... + \sqrt {2017} - \sqrt {2013} \\
= 1 + \sqrt 5 - \sqrt 5 + \sqrt 9 - \sqrt 9 + ... + \sqrt {2013} - \sqrt {2013} + \sqrt {2017} \\
= 1 + \sqrt {2017} \\
\Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]
14 tháng 11 2015
Viết ngược lại đề bài nha rồi trục căn thức
\(2S=\sqrt{2013}-\sqrt{2011}+\sqrt{2011}-\sqrt{2009}+....+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}\)
\(S=\frac{\sqrt{2013}-\sqrt{3}}{2}\)
6 tháng 10 2019
Trục căn thức:
\(C=\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\)
\(+\frac{\left(\sqrt{2017}-\sqrt{2015}\right)}{\left(\sqrt{2017}+\sqrt{2015}\right)\left(\sqrt{2017}-\sqrt{2015}\right)}\)
\(C=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+...+\frac{\sqrt{2017}-\sqrt{2015}}{2017-2015}\)
\(C=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+...+\frac{\sqrt{2017}-\sqrt{2015}}{2}\)
\(C=\frac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2015}}{2}\)
\(C=\frac{\sqrt{2017}-1}{2}\)
TB
31 tháng 7 2018
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{2018}-\sqrt{2019}}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{2019}-\sqrt{2018}}{2019-2018}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}\)
\(=\sqrt{2019}-\sqrt{2}\)
6 tháng 8 2016
Xét biểu thức phụ : \(\frac{1}{\left(2n+3\right)\sqrt{2n+1}+\left(2n+1\right)\sqrt{2n+3}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2\sqrt{2n+1}.\sqrt{2n+3}}=\frac{1}{2}\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\ge1\)
Áp dụng : \(S=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{103}}\right)\)
Bạn ơi cái này mk chỉ ghi cách làm và ct thôi nha
đây dùng hàng đẳng thức (a-b)(a+b)=a^2-b^2
còn kia là công thức toán lớp 6
\(\frac{1}{\sqrt{3}+\sqrt{1}}=\frac{\sqrt{3}-\sqrt{1}}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}=\frac{\sqrt{3}-\sqrt{1}}{\sqrt{3^2}-\sqrt{1^2}}=\frac{1}{2}\left(\sqrt{3}-\sqrt{1}\right)\)
Tương tự:
\(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{1}{2}\left(\sqrt{5}-\sqrt{3}\right)\)
.....
\(\frac{1}{\sqrt{2019}+\sqrt{2017}}=\frac{1}{2}\left(\sqrt{2019}-\sqrt{2017}\right)\)
Cộng các vế với nhau ta được:
\(S=\frac{1}{2}\left(\sqrt{2019}-\sqrt{1}\right)=\frac{1}{2}\left(\sqrt{2019}-1\right)\)