\(Q=\frac{1}{\sqrt{1}+\sqrt{5}}\)+ \(\frac{1}{\sqrt{5}+\sqrt{9}}\)+\(\frac{1}{\sqrt{9}+\sqrt{13}}\)+...+\(\frac{1}{\sqrt{2013}+\sqrt{2017}}\) =?
tính \(\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+\frac{1}{9\sqrt{13}+13\sqrt{9}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)
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
\(\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)
\(=\frac{1}{4}.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2013}}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{\sqrt{2013}}\right)\)
Giải chi tiết hộ mk
Tính các tổng sau:
a)\(T=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2013}+\sqrt{2017}}\)
b)\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
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
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5 + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013} + \sqrt {2017} }}\\
= \frac{{(\sqrt 5 + 1)(\sqrt 5 - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017} + \sqrt {2013} )(\sqrt {2017} - \sqrt {2013} )}}{{\sqrt {2013} + \sqrt {2017} }}\\
= \sqrt 5 - 1 + \sqrt 9 - \sqrt 5 + ... + \sqrt {2017} - \sqrt {2013} \\
= 1 + \sqrt 5 - \sqrt 5 + \sqrt 9 - \sqrt 9 + ... + \sqrt {2013} - \sqrt {2013} + \sqrt {2017} \\
= 1 + \sqrt {2017} \\
\Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]
tính
A=\(\frac{1}{\sqrt{1}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2013}+\sqrt{2017}}\)
\(A=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+...+\frac{\sqrt{2017}-\sqrt{2013}}{4}\)
\(A=\frac{\sqrt{2017}-1}{4}\)
Tính \(A=\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)
\(\frac{1}{n\sqrt{n+4}+\left(n+4\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+4\right)}.\left(\sqrt{n}+\sqrt{n+4}\right)}=\frac{\sqrt{n+4}-\sqrt{n}}{4.\sqrt{n\left(n+4\right)}}=\frac{1}{4}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+4}}\right)\)
Áp dụng công thức trên ta có:
\(A=\frac{1}{4}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2015}}\right)=\frac{1}{4}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2015}}\right)=\frac{\sqrt{2015}-1}{4\sqrt{2015}}\)
Rút gọn: \(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2014}+\sqrt{2018}}\)
\(A=\frac{1}{\sqrt{1}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2014}+\sqrt{2018}}\)
\(\Rightarrow A=\sqrt{5}-\sqrt{1}+\sqrt{9}-\sqrt{5}+...+\sqrt{2018}-\sqrt{2014}\)
\(\Rightarrow A=-\sqrt{1}+\sqrt{2018}\)
cho mk nha
Ai trên 11 điểm cho mình nha câu dưới 3 mk lại
Bạn ơi trục căn thức sao không còn mẫu vậy
trục căn thức k còn mẫu vì bn ý copy bn ẹ, copy của Thắng Nguyễn :))
Rút gọn biểu thức:
\(Q=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+.....+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(Q=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
=> \(Q=\frac{1-\sqrt{5}}{-4}+\frac{\sqrt{5}-\sqrt{9}}{-4}+\frac{\sqrt{9}-\sqrt{13}}{-4}+...+\frac{\sqrt{2001}-\sqrt{2005}}{-4}\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}\right)\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{2005}\right)\)
=> \(Q=\frac{\sqrt{2005}-1}{4}\)
Rút gọn
A= \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
B= \(\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{2}}\right)\left(4-\sqrt{10+2\sqrt{2}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5.1}+1}=8+2\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(B=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=-\frac{1}{4}\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+....+\sqrt{2001}-\sqrt{2005}\right)\)
\(=-\frac{1}{4}\left(1-\sqrt{2005}\right)\)
\(=10,94430659\)
\(\text{Lm hơi vắn tắt thông cảm nha!!}\)
Tính :
a ) \(S=\frac{1}{\sqrt{1}\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+.....+\)\(\frac{1}{\sqrt{2017}+\sqrt{2019}}\)
b ) \(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+....+\frac{1}{\sqrt{100}+\sqrt{102}}\)
c ) \(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{101}}\)
d ) \(S=\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}}+....+\frac{1}{\sqrt{2016}+\sqrt{2019}}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
Rút gọn biểu thức:
\(\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2009}+\sqrt{2013}}\)
\(\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2009}+\sqrt{2013}}\)
=\(\frac{-1+\sqrt{5}+3-\sqrt{5}-3+\sqrt{13}+...+\sqrt{2013}-\sqrt{2009}}{4}\)
=\(\frac{-1-\sqrt{2009}}{4}\)
=\(-\frac{1+7\sqrt{41}}{4}\)