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\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+........+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=\frac{1}{\sqrt{1}\sqrt{2}\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{\sqrt{2}\sqrt{3}\left(\sqrt{2}+\sqrt{3}\right)}+........+\frac{1}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2009}+\sqrt{2010}\right)}\)
\(=\frac{\left(\sqrt{2010}-\sqrt{2009}\right)\left(\sqrt{2010}+\sqrt{2009}\right)}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2010}+\sqrt{2009}\right)}+.......+\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}=1-\frac{1}{\sqrt{2010}}=1-\frac{\sqrt{2010}}{2010}\)
Đặt B = \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{50}}\)
= \(1+2\left(\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+...+\dfrac{1}{2\sqrt{50}}\right)\)
Đặt \(A=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+...+\dfrac{1}{2\sqrt{50}}\)
Xét A < \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{49}+\sqrt{50}}\)
=> A < \(\dfrac{\sqrt{2}-\sqrt{1}}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+...+\dfrac{\sqrt{50}-\sqrt{40}}{1}\)
=> A < -1 + \(\sqrt{50}\)
=> 2A < -2 + \(10\sqrt{2}\)
=> 2A + 1 = B < -2 + \(10\sqrt{2}\) + 1
=> B < -1 + \(10\sqrt{2}\) < \(10\sqrt{2}\) (1)
Xét \(\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)
=> \(\dfrac{1}{\sqrt{1}}>2\left(\sqrt{2}-\sqrt{1}\right)\)
\(\dfrac{1}{\sqrt{2}}>2\left(\sqrt{3}-\sqrt{2}\right)\)
\(\dfrac{1}{\sqrt{3}}>2\left(\sqrt{4}-\sqrt{3}\right)\)
...
\(\dfrac{1}{\sqrt{50}}>2\left(\sqrt{51}-\sqrt{50}\right)\)
=> B > 2(\(\sqrt{51}-\sqrt{1}\))
=> B >-2 + \(10\sqrt{2}\) > \(5\sqrt{2}\)
Cảm ơn bạn nha. Mà bạn bị nhầm 49 thành 40 ở dòng thứ 5 đó.
\(B=\sqrt{1+2008^2+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}=\sqrt{\dfrac{2009^2+2008^2.2009^2+2008^2}{2009^2}}+\dfrac{2008}{2009}=\dfrac{\sqrt{2009^2+\left(2009-1\right)^2.2009^2+2008^2}}{2009}+\dfrac{2008}{2009}=\dfrac{\sqrt{2009^2+2009^4-2.2009.2009^2+2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4+2.2009^2-2.\left(2008+1\right).2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4+2.2009^2-2.2008.2009^2-2.2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4-2.2008.2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{\left(2009^2-2008\right)^2}+2008}{2009}=\dfrac{2009^2-2008+2008}{2009}=2009\in N\)
Vậy B có giá trị là một số tự nhiên
Xét các số thực a, b, c thỏa mãn \(a+b+c=0\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2.\frac{a+b+c}{abc}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
Ta có:
\(B=\sqrt{1+2008^2+\frac{2008^2}{2009^2}}+\frac{2008}{2009}\)
\(=\sqrt{2008^2}.\sqrt{\frac{1}{2018^2}+\frac{1}{1^2}+\frac{1}{2009^2}}+\frac{2008}{2009}\)
\(=2008.\sqrt{\frac{1}{2018^2}+\frac{1}{1^2}+\frac{1}{\left(-2009\right)^2}}+\frac{2008}{2009}\)
\(=2008.\left|\frac{1}{2008}+1-\frac{1}{2009}\right|+\frac{2008}{2009}\)
\(=2008.\left(\frac{1}{2008}+1-\frac{1}{2009}\right)+\frac{2008}{2009}\)
\(=2008.\left(\frac{1}{2008}+1-\frac{1}{2009}+\frac{1}{2009}\right)\)
\(=2008.\frac{2009}{2008}=2009\in\text{N}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{2008}+\sqrt{2009}}=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+\dfrac{\sqrt{4}-\sqrt{3}}{4-3}+...+\dfrac{\sqrt{2009}-\sqrt{2008}}{2009-2008}=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2009}-\sqrt{2008}=\sqrt{2009}-\sqrt{2}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{2008}+\sqrt{2009}}\)
\(=\dfrac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}+...+\dfrac{\sqrt{2008}-\sqrt{2009}}{\left(\sqrt{2008}+\sqrt{2009}\right)\left(\sqrt{2008}-\sqrt{2009}\right)}\)
\(=\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+...+\dfrac{\sqrt{2008}-\sqrt{2009}}{2008-2009}\)
\(=-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{2008}+\sqrt{2009}\)
\(=-\sqrt{2}+\sqrt{2009}\)
2/ \(\sqrt{4+\sqrt{4+...+\sqrt{4}}}< \sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{7+\sqrt{4}}}}}=3\)
1/ Ta có:
\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\left(\dfrac{n^2+n+1}{n\left(n+1\right)}\right)^2}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow C=99+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=100-\dfrac{1}{100}=\dfrac{9999}{100}\)
Ta có các số trong dãy đều có dạng 1/[ (n + 1)√n ]
Ta có: 1/[ (n + 1)√n ] = (√n)/[ (n + 1)√n.√n ] = (√n)/[ (n + 1)n ] = (√n).1/[ (n + 1)n ]
Do 1/[ (n + 1)n ] = 1/n - 1/(n + 1) (mình nghĩ bạn biết cái này)
=> (√n).1/[ (n + 1)n ] = (√n).[ 1/n - 1/(n + 1) ]
Ta có 1/n - 1/(n + 1) = (1/√n)² - [ 1/√(n + 1) ]²
= [ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ]
=> 1/n - 1/(n + 1) = [ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ]
=> (√n).[ 1/n - 1/(n + 1) ] = (√n).[ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ]
Nhân √n với [ 1/√n + 1/√(n + 1) ] ta được
(√n).[ 1/√n + 1/√(n + 1) ]. [ 1/√n - 1/√(n + 1) ] = [ 1 + (√n)/√(n + 1) ].[ 1/√n - 1/√(n + 1) ]
=> 1/[ (n + 1)√n ] = [ 1 + (√n)/√(n + 1) ].[ 1/√n - 1/√(n + 1) ] (1)
Do (√n)/√(n + 1) < √(n + 1)/√(n + 1)
=> (√n)/√(n + 1) < 1
=> 1 + (√n)/√(n + 1) < 1 + 1
=> 1 + (√n)/√(n + 1) < 2
=> [ 1 + (√n)/√(n + 1) ].[ 1/√n - 1/√(n + 1) ] < 2.[ 1/√n - 1/√(n + 1) ] (2)
Từ (1) và (2) => 1/[ (n + 1)√n ] < 2.[ 1/√n - 1/√(n + 1) ]
Áp dụng ta được
1/2 < 2( 1 - 1/√2)
1/3√2 < 2(1/√2 - 1/√3)
....
1/(n+1)√n < 2(1/√n - 1/√(n + 1) )
=> 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2( 1 - 1/√2) + 2(1/√2 - 1/√3) + ... + 2(1/√n - 1/√(n + 1) )
=> 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2( 1 - 1/√2 + 1/√2 - 1/√3 + ... + 1/√n - 1/√(n + 1) )
=> 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2(1 - 1/√(n + 1) ) (3)
Do 1√(n + 1) > 0
=> -1√(n + 1) < 0
=> 1 -1√(n + 1) < 1
=> 2(1 - 1/√(n + 1) ) < 2 (4)
Từ (3) và (4) => 1/2 + 1/3√2 + 1/4√3 +.....+ 1/(n+1)√n < 2
\(\dfrac{1}{\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n\left(n+1\right)}}.\dfrac{1}{\sqrt{n+1}}\) . Do \(\sqrt{n+1}>\dfrac{\sqrt{n}+\sqrt{n+1}}{2}\)
\(\Rightarrow\dfrac{1}{\sqrt{n\left(n+1\right)}}.\dfrac{1}{\sqrt{n+1}}< \dfrac{1}{\sqrt{n\left(n+1\right)}}.\dfrac{2}{\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Vậy \(\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán:
\(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{2009\sqrt{2008}}< 2\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2008}}-\dfrac{1}{\sqrt{2009}}\right)\)
\(\Rightarrow VT< 2\left(1-\dfrac{1}{\sqrt{2009}}\right)< 2-\dfrac{2}{\sqrt{2009}}< 2\) (đpcm)