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\(\sqrt{2+\sqrt{2+...+\sqrt{2}}}< \sqrt{2+\sqrt{2+...+\sqrt{4}}}=2\)
Vậy \(\sqrt{2+...+\sqrt{2}}< 2\)
a) Ta có : \(5>2\Rightarrow\sqrt{5}>\sqrt{2}\)
b) Vì \(8>5\Rightarrow\sqrt{8}>\sqrt{5}\Rightarrow2\sqrt{2}>5\)
c) VÌ \(-32>-45\Rightarrow-\sqrt{32}>-\sqrt{45}\Rightarrow-4\sqrt{2}>-\sqrt{5}\)
d) Vì \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Leftrightarrow2\sqrt{3}< 3\sqrt{2}\)
Ta có:
\(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}=\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)^2}{\left(1+\sqrt{n}+\sqrt{n+1}\right)\left(1-\sqrt{n}+\sqrt{n+1}\right)}=\frac{2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}}{2\left(1+\sqrt{n+1}\right)}\)
\(=\frac{\left[2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}\right]\left(1-\sqrt{n+1}\right)}{2\left(1+\sqrt{n+1}\right)\left(1-\sqrt{n+1}\right)}=\frac{-2n\sqrt{n+1}+2n\sqrt{n}}{-2n}=\sqrt{n+1}-\sqrt{n}\)
Suy ra:
\(Q=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2016}=\sqrt{2017}-\sqrt{2}< \sqrt{2017}-1=R\)
Vậy Q < R.
Lời giải:
\(\sqrt{20152015}+\sqrt{20152017}-2\sqrt{20152016}=(\sqrt{20152015}-\sqrt{20152016})+(\sqrt{20152017}-\sqrt{20152016})\)
\(=\frac{-1}{\sqrt{20152015}+\sqrt{20152016}}+\frac{1}{\sqrt{20152017}+\sqrt{20152016}}\)
Dễ thấy: $0< \sqrt{20152015}+\sqrt{20152016}<\sqrt{20152017}+\sqrt{20152016}}$
$\Rightarrow \frac{1}{\sqrt{20152015}+\sqrt{20152016}}>\frac{1}{\sqrt{20152017}+\sqrt{20152016}}$
$\Rightarrow \frac{-1}{\sqrt{20152015}+\sqrt{20152016}}+\frac{1}{\sqrt{20152017}+\sqrt{20152016}}< 0$
$\Rightarrow \sqrt{20152015}+\sqrt{20152017}< 2\sqrt{20152016}$
Lời giải:
Ta có:
$\sqrt{2015.2015}+\sqrt{2015.2017}=\sqrt{2015}(\sqrt{2015}+\sqrt{2017})$
Mà:
$(\sqrt{2015}+\sqrt{2017})^2=4032+2\sqrt{2015.2017}$
$=4032+2\sqrt{(2016-1)(2016+1)}=4032+2\sqrt{2016^2-1}$
$< 4032+2\sqrt{2016^2}=4.2016$
$\Rightarrow \sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}$
$\Rightarrow \sqrt{2015.2015}+\sqrt{2015.2017}=\sqrt{2015}(\sqrt{2015}+\sqrt{2017})< \sqrt{2015}.2\sqrt{2016}$
Vậy......