Ta có:

\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\\\dfrac{1}{\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\end{matrix}\right.\)

Thế vô giải tiếp

\(\frac{1}{(n+1)\sqrt{n} }=\frac{\sqrt{n} }{n(n+1)}=\sqrt{n} (\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } )(\frac{1}{\sqrt{n} } +\frac{1}{\sqrt{n+1} } )=(1+\frac{\sqrt{n} }{\sqrt{n+1} } )(\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } <2(\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } )\)

Áp dụng BĐT vừa CM ta có

A< 2(1-\(\frac{1}{\sqrt{2} } +\frac{1}{\sqrt{2} } -\frac{1}{\sqrt{3} } +...+\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } \))<2(đpcm)

Chứng minh biểu thức đó <2

Với mọi \(n\in N^{\cdot}\), ta có

\(\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Leftrightarrow1< 2\left(n+1\right).\sqrt{n}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Leftrightarrow0< n+1-2\sqrt{n+1}.\sqrt{n}+n\)

\(\Leftrightarrow0< \left(\sqrt{n+1}-\sqrt{n}\right)^2\)(Luôn đúng vì n thuộc N^*)

Do đó: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...\dfrac{1}{2005\sqrt{2004}}< 2\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2004}}-\dfrac{1}{\sqrt{2005}}\right)\)

\(=2\left(1-\dfrac{1}{\sqrt{2005}}\right)< 2\)

\(b,\) Ta có:

\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)

Thay:

\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)

\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)

\(...\)

\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)

Tiếp phần b ( do máy lag) :3

Cộng 2 vế với nhau, ta có:

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\\ =1-\dfrac{1}{\sqrt{2007}}\)

a) A=\(\dfrac{1}{\sqrt{3}+\sqrt{5}}\)+\(\dfrac{1}{\sqrt{5}+\sqrt{7}}\)+\(\dfrac{1}{\sqrt{7}+\sqrt{9}}\)+...+\(\dfrac{1}{\sqrt{97}+\sqrt{99}}\)

=\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)+\(\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)+\(\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\left(\sqrt{9}-\sqrt{7}\right)}\)+...+\(\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}\)

=\(\dfrac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}}{2}\)

=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)

vậy A=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)

Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)

\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)

\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)

\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)

\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)

\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)

=>A là số hữu tỉ (ĐPCM)