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\(U\left(n\right)=\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}\)
\(U\left(n\right)=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n.\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(U\left(n\right)=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}\sqrt{n+1}\right)^2}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)
\(a)\) \(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=a-b\)
\(b)\) \(B=a-b=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)\(\Rightarrow\)\(B^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(B^2=4-2\sqrt{4-3}=4-2=2\)\(\Rightarrow\)\(B=\sqrt{2}\) ( vì \(B>0\) )
...
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
\(\Rightarrow\frac{B}{2}=\frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{2}}+...+\frac{1}{2\sqrt{2010}}\)
\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2010}+\sqrt{2011}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2011}-\sqrt{2010}}{2011-2010}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\)
\(=\sqrt{2011}-1>43\)
=>B> 43.2=86
Vậy B> 86
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TA có \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}}>\frac{2}{\sqrt{n+1}+\sqrt{n}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\)
Áp dụng BĐT ta có :
\(B=\frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{2010}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)=2\left(\sqrt{2011}-1\right)\) (1)
\(2\left(\sqrt{2011}-1\right)>2\left(\sqrt{1936}-1\right)=2\left(44-1\right)=86\) (2)
Từ (1) và (2) => B > 86