Ta có:

\(B=20152015.20152017=\left(20152016-1\right)\left(20152016+1\right)=20152016^2-1\)

Lại có,  \(A=20152016^2\)

Vậy,   \(A>B\)

\(B=20152015.20152017=\left(20152016-1\right)\left(20152016+1\right)\)

\(B=\left(20152016-1\right)\left(20152016+1\right)=20152016^2-1< A\)

A>B

Bài 18:

Ta có:

\(2015^{2015}-2015^{2014}=2015^{2014}\cdot\left(2015-1\right)=2015^{2014}\cdot2014\)

\(2015^{2016}-2015^{2015}=2015^{2015}\cdot\left(2015-1\right)=2015^{2015}\cdot2014\)

Mà: \(2014< 2015\)

\(\Rightarrow2015^{2014}< 2015^{2015}\)

\(\Rightarrow2015^{2014}\cdot2014< 2015^{2015}\cdot2014\)

\(\Rightarrow2015^{2015}-2015^{2014}< 2015^{2016}-2015^{2015}\)

Vậy: ... 

6 : (x-2)

Ta thấy:

A = \(\frac{20162017}{20162016}\) và     B =  \(\frac{20152016}{20152015}\)

A  =  \(\frac{20162016}{20162016}\)+  \(\frac{1}{20162016}\)  =   \(1\) +   \(\frac{1}{20162016}\)

B  =   \(\frac{20152015}{20152015}\) +   \(\frac{1}{20152015}\)=   \(1\)  +    \(\frac{1}{20152015}\)

Vì:     \(\frac{1}{20162016}\)   \(< \)       \(\frac{1}{20152015}\)

Nên:    \(A\)    \(< \)    \(B\)

~ HokT~

A>b mình  nghĩ vậy 

Ta có: \(\sqrt{1+\sqrt{2+\sqrt{3}}}=\sqrt{1+\sqrt{\dfrac{4+2\sqrt{3}}{2}}}=\sqrt{1+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}}\)

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

\(\Rightarrow\) cần so sánh \(\sqrt{\dfrac{\sqrt{2}+\sqrt{3}+1}{\sqrt{2}}}\) với 2

Bình phương 2 vế (cả 2 vế đề không âm nên bình phương được)

\(\Rightarrow\) cần so sánh \(\dfrac{\sqrt{2}+\sqrt{3}+1}{\sqrt{2}}\) với 4

\(\Rightarrow\) cần so sánh \(\sqrt{2}+\sqrt{3}+1\) với \(4\sqrt{2}\)

\(\Rightarrow\) cần so sánh \(\sqrt{3}+1\) với \(3\sqrt{2}\)

Ta có; \(3\sqrt{2}=2\sqrt{2}+\sqrt{2}=\sqrt{8}+\sqrt{2}\)

Vì \(\left\{{}\begin{matrix}8>3\\2>1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{8}>\sqrt{3}\\\sqrt{2}>\sqrt{1}\end{matrix}\right.\Rightarrow\sqrt{8}+\sqrt{2}>\sqrt{3}+1\)

\(\Rightarrow\sqrt{1+\sqrt{2+\sqrt{3}}}< 2\)

\(\frac{\sqrt{21} - \sqrt{13}}{35 - 2\sqrt{273}} = \frac{\sqrt{21} - \sqrt{13}}{(\sqrt{21} - \sqrt{13})^2} = \frac{1}{\sqrt{21} - \sqrt{13}}\)

\(=\frac{\sqrt{21}+\sqrt{13}}{8}>\frac{\sqrt{16}+\sqrt9}{8}=\frac{4+3}{8}=\frac78\)

\(\frac{\sqrt{10} - \sqrt{5}}{16 - 10\sqrt{2}}=\frac{\sqrt{5}(\sqrt{2} - 1)}{16 - 10\sqrt{2}}=\frac{\sqrt{5}}{6\sqrt{2} - 4}=\frac{\sqrt5}{2\left(3\sqrt2-2\right)}\)

\(\frac{\sqrt5}{2\left(3\sqrt2-2\right)}-\frac18=\frac{8\sqrt5-6\sqrt2+4}{16\left(3\sqrt2-2\right)}=\frac{\sqrt{320}-\sqrt{72}+4}{16\left(3\sqrt2-2\right)}>\frac{17-8+4}{16\left(3\sqrt2-2\right)}=\frac{13}{16\left(3\sqrt2-2\right)}>5>\frac18\)

Do đó: \(\frac{\sqrt{21}-\sqrt{13}}{35-2\sqrt{273}}+\frac{\sqrt{10}-\sqrt5}{16-10\sqrt2}\) >1(ĐPCM)

\(\sqrt{2+\sqrt{2+...+\sqrt{2}}}< \sqrt{2+\sqrt{2+...+\sqrt{4}}}=2\)

Vậy \(\sqrt{2+...+\sqrt{2}}< 2\)

thừa số 2 sau cùng

\(\Leftrightarrow B^2=\dfrac{x}{\left(\sqrt{x}-2\right)^2}>0\)

\(\Leftrightarrow B>\sqrt{B}\)