cho \(S=1+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{5}}+...+\dfrac{1}{\sqrt{99}}\) . CMR: S > 9
Tính các tổng sau:
\(T=\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}+......+\dfrac{1}{\sqrt{2013}+\sqrt{2017}}}\)
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
CMR:\(\dfrac{1}{1+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{9}+\sqrt{11}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)
1. Tìm số tự nhiên n sao cho :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{n.\left(n+1\right)}=\dfrac{2999}{3000}\)
2. Tính :
a ) \(S=2018.3+2018.4+2018.5+...+2018.2018\)
b ) \(\dfrac{1}{\sqrt{8}+\sqrt{10}}+\dfrac{1}{\sqrt{10}+\sqrt{12}}+\dfrac{1}{\sqrt{12}+\sqrt{14}}+...+\dfrac{1}{\sqrt{200}+\sqrt{202}}\)
c ) \(S=5.21^2+5.21^3+5.21^4+....+5.21^{2018}\)
d ) \(A=9+99+999+9999+...+9..9\)( 99 chữ số 9)
e ) 72+772+7772+...+77...72( 77 chữ số 7 )
2. Tính tổng :
a ) \(S=\dfrac{1}{3\sqrt{1}+3\sqrt{3}}+\dfrac{1}{3\sqrt{3}+3\sqrt{5}}+...+\dfrac{1}{3\sqrt{2017}+3\sqrt{2019}}\)
b ) S = \(\dfrac{1}{\sqrt{2.2}+\sqrt{2.3}}+\dfrac{1}{\sqrt{2.3}+\sqrt{2.4}}+\dfrac{1}{\sqrt{2.4}+\sqrt{2.5}}+...+\dfrac{1}{\sqrt{2.2018}+\sqrt{2.2019}}\)
Câu 1:
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
=>n+1=3000
hay n=2999
Tính :
a ) S= 5+55+555+...+55...5 ( 50 chữ số 5 )
b ) S= 75+755+7555+...+755...5 ( 50 chữ số 5 )
c ) \(S=\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2017} +\sqrt{2019}}\)
d ) \(S=\dfrac{1}{\sqrt{3}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{12}}+...+\dfrac{1}{\sqrt{2016}+\sqrt{2019}}\)
Bài 1: Cho A = \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
So sánh A với 1
Bài 2: Tính
A = \(\left(\dfrac{3}{\sqrt{2}+1}+\dfrac{14}{2\sqrt{2}-1}-\dfrac{4}{2-\sqrt{2}}\right).\left(\sqrt{8}+2\right)\)
Bài 3: Tính tổng
S=\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2018}+\sqrt{2019}}\)
Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
Tính tổng sau: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
S=\(\dfrac{1}{1\sqrt{2}+2\sqrt{2}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+\dfrac{1}{3\sqrt{4}+4\sqrt{3}}+....+\dfrac{1}{99\sqrt{100}+100\sqrt{99}}\)
So sánh A với 1
\(A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có :
\(A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+.....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}+........+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\) \(=1-\dfrac{1}{\sqrt{100}}< 1\)
Vậy \(A< 1\)
Tính:
1) \(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}\)
2) \(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)
3) \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
4) \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{\sqrt{5}-3}\)
5) \(\dfrac{1}{\sqrt{2}-\sqrt{6}}-\dfrac{1}{\sqrt{6}+\sqrt{2}}\)
LM CHI TIẾT GIÚP MK NHÉ
4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}-3-\sqrt{5}}{4}\)
\(=\dfrac{-\sqrt{5}}{2}\)