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\(Tongquat:\)
\(\sqrt{1+\frac{1}{n}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n}+\frac{2}{n}-\frac{2}{n+1}-\frac{2}{n\left(n+1\right)}+\frac{1}{\left(n+1\right)^2}}\)
\(=\sqrt{\left(1+\frac{1}{n}\right)^2-2\left(1+\frac{1}{n}\right)\frac{1}{n+1}+\frac{1}{n+1}}=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}\)
\(=|1+\frac{1}{n}-\frac{1}{n+1}|=1+\frac{1}{n}-\frac{1}{n+1}\)
Thay vào ta có:
\(P=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+.........-\frac{1}{2017}\)
\(P=2015+\frac{1}{2}-\frac{1}{2017}=2015+\frac{2015}{4034}\)
\(\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)\)
Với n thuộc N*, ta có:
\(\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{2\left(n+1-n-1\right)}{n\left(n+1\right)}}\)\)
\(\(=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2.1.\frac{1}{n}-2.1.\frac{1}{n+1}-2.\frac{1}{n}.\frac{1}{\left(n+1\right)}}\)\)
\(\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n-1}\right)^2}=1+\frac{1}{n}-\frac{1}{n-1}\)\). Áp dụng vô bài, ta có:
\(\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)\)
\(\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)\)
\(\(=2016+\frac{1}{2}-\frac{1}{2018}=2016\frac{504}{1009}\)\)
P/s: Lại là thằng quỷ Thắng
Xét số hạng tổng quát
\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=1^2+\left(\frac{1}{k}\right)^2+\left(\frac{1}{k+1}\right)^2+2.1.\frac{1}{k}-2.\left(\frac{1}{k}.\frac{1}{k+1}\right)-2.1.\frac{1}{k+1}\)
\(=\left(1+\frac{1}{k}-\frac{1}{k+1}\right)^2\)
( Vì \(\frac{1}{k}-\frac{1}{k\left(k+1\right)}-\frac{1}{k+1}=\frac{k+1-1-k}{k\left(k+1\right)}=0\) )
Vậy thì \(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=1+\frac{1}{k}-\frac{1}{k+1}\)
Vậy \(A=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)
\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)
\(=2016+\frac{1}{2}-\frac{1}{2018}=2016\frac{504}{1009}\)
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
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) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
VT<1/(3^2-1)+1/(5^2-1)+...+1/(2017^2-1)=1/(2.4)+1/(4.6)+...+1/(2016.2018)
=1/2 . (1/2-1/4+1/4-1/6+...+1/2016-1/2018)=1/4-1/(2.2018)<1/4