Chứng minh rằng: A=\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}\)là số hữu tỉ
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Xét : \(\left(\frac{1}{k-1}-\frac{1}{k}+1\right)^2=\frac{1}{k^2}+\frac{1}{\left(k-1\right)^2}+1+2\left(-\frac{1}{k\left(k-1\right)}-\frac{1}{k}+\frac{1}{k-1}\right)\)
\(=\frac{1}{k^2}+\frac{1}{\left(k-1\right)^2}+1\)
\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=\left|\frac{1}{k-1}-\frac{1}{k}+1\right|\)với k thuộc N* , k > 1
Áp dụng : \(\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{1999^2}+\frac{1}{2000^2}}\)
\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{1999}-\frac{1}{2000}\right)\)
\(=1998+\frac{1}{2}+-\frac{1}{2000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{n\sqrt{\left(n+1\right)}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{\left(n+1\right)}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào ta được
\(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{99\sqrt{100}+100\sqrt{99}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta sẽ chứng minh với \(a\in Q\) thì \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\) là số hữ tỉ
ta có \(M=\frac{1}{1}+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{1}{1}+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}+\frac{2}{a}-\frac{2}{a+1}-\frac{2}{a\left(a+1\right)}-\frac{2}{a}+\frac{2}{a+1}+\frac{2}{a\left(a+1\right)}\)
\(=\left(\frac{1}{1}+\frac{1}{a}-\frac{1}{a+1}\right)^2+2\left(\frac{1}{a}+\frac{1}{a\left(a+1\right)}-\frac{1}{a+1}\right)\)
\(=\left(1+\frac{1}{a}+\frac{1}{a+1}\right)^2+2\left(\frac{1+a-\left(a+1\right)}{a\left(a+1\right).1}\right)=\left(1+\frac{1}{a}+\frac{1}{a+1}\right)^2\)
=> \(\sqrt{M}=\left|1+\frac{1}{a}+\frac{1}{a+1}\right|\) là số hữu tỉ
=> A lá số hữ tỉ
Áp dụng thì ta có mỗi phân thức là số hữ tỉ nên tổng của nó là sô hưux tỉ
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)
Với a , b , c là số hữu tỉ t/m a = b + c ta luôn có \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right|\in Q\)
Thật vậy : \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2-2\left(\frac{1}{bc}-\frac{1}{ac}-\frac{1}{ab}\right)}\)
\(=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2-\frac{2.abc\left(a-b-c\right)}{a^2b^2c^2}}\)(quy đồng lên )
\(=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2}\left(\text{do a-b-c=0}\right)\)
\(=\left|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right|\in Q\)
Áp dụng ta được \(A=\left|\frac{1}{3}-\frac{1}{2}-1\right|+\left|\frac{1}{4}-\frac{1}{3}-1\right|+...+\left|\frac{1}{2000}-\frac{1}{1999}-1\right|\)là số hữu tỉ
Vậy A là số hữu tỉ