Chứng minh: \(\frac{1}{2\sqrt{2}+1\sqrt{1}}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< 1-\frac{1}{\sqrt{n+1}}\)
Chứng minh:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< 1-\frac{1}{\sqrt{n+1}}\left(n\in N\right)\)
Xét dạng tổng quát có: \(\frac{1}{\sqrt{n+1}\left(n+1\right)+n\sqrt{n}}=\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)-\sqrt{n\left(n+1\right)}}\)
\(< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}-\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta có:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< 1-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
.....
\(\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cộng vế theo vế =>\(VT< 1-\frac{1}{\sqrt{n+1}}\left(ĐPCM\right)\)
a/Chứng minh rằng \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b/Áp dụng chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
chứng minh : \(\frac{1}{2\sqrt{2}+1\sqrt{1}}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< 1-\frac{1}{\sqrt{n+1}}\)
Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)
a, Chứng minh
\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, Áp dụng
\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}......+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
a)\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2.n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b)\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\( S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
\(a,\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(n+1-n\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{\sqrt{n-1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
1) Chứng minh: \(2\sqrt{n}-3< \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}-2\forall n\ge2\)
2) Thu gọn: \(A=5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
Chứng minh rằng \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\) (với \(n\in N^{\cdot}\))
Áp dụng cho S = \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
Chứng minh 18<S<19 ?
Lời giải:
Liên hợp ta thấy:
\(2(\sqrt{n+1}-\sqrt{n})=2.\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{2}{\sqrt{n+1}+\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(1)\)
\(2(\sqrt{n}-\sqrt{n-1})=2.\frac{n-(n-1)}{\sqrt{n}+\sqrt{n-1}}=\frac{2}{\sqrt{n}+\sqrt{n-1}}>\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(2)\)
Từ \((1);(2)\Rightarrow 2(\sqrt{n+1}-\sqrt{n})< \frac{1}{\sqrt{n}}< 2(\sqrt{n}-\sqrt{n-1})\)
------------------------
Áp dụng vào bài toán:
\(S=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>1+2(\sqrt{3}-\sqrt{2})+2(\sqrt{4}-\sqrt{3})+...+2(\sqrt{101}-\sqrt{100})\)
\(\Leftrightarrow S>1+2(\sqrt{101}-\sqrt{2})>18(*)\)
Và:
\(S< 1+2(\sqrt{2}-\sqrt{1})+2(\sqrt{3}-\sqrt{2})+....+2(\sqrt{100}-\sqrt{99})\)
\(\Leftrightarrow S< 1+2(\sqrt{100}-\sqrt{1})=19(**)\)
Từ $(*); (**)$ suy ra $18< S< 19$ (đpcm)
a. Chứng minh : \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. Áp dụng : Tính giá trị của biểu thức :
\(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
cảm ơn các bạn trước nhé!
chứng minh \(\frac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}< \frac{1}{2}\)
Đặt biểu thức trên là A.
Ta có: \(\left(\sqrt{n+1}+\sqrt{n}\right)\)).(\(\sqrt{n+1}-\sqrt{n}\))=1
=>\(\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right)}=\left(\sqrt{n+1}-\sqrt{n}\right)\)
Từ trên: \(\frac{1}{\left(2n+1\right).\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{2n+1}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1+n}\)
Lại có :\(\frac{\sqrt{n+1}-\sqrt{n}}{\left(n+1\right)+n}< \frac{1}{2}.\frac{\sqrt{n+1}-\sqrt{n}}{\left(n+1\right).n}=\frac{1}{2}.\left(\frac{1}{n}-\frac{1}{n+1}\right)\)(Bất đẳng thức Cô-si)
Thế số vào, ta được :
A<\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...-\frac{1}{\sqrt{n+1}}\right)\)=\(\frac{1}{2}.\left(1-\frac{1}{\sqrt{n+1}}\right)< \frac{1}{2}\)