ồ, lâu h ms gặp

a,

Dễ thấy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)

Áp dụng khi \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)

Ta có:

\(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2016}+1+\left(2005^2-1\right)}{2005^{2017}+1+\left(2005^2-1\right)}=\dfrac{2005^{2016}+2005^2}{2005^{2017}+2005^2}=\dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)

Vậy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)

b,

\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)

Vì \(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)

c,

\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)

Vì \(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)

Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)

Áp dụng tính chất:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)

\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)

\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)

\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)

\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)

\(B< A\)

Ta thấy: \(\left\{{}\begin{matrix}A=\dfrac{2005^{2014}+1}{2005^{2015}+1}< 1\\B=\dfrac{2005^{2015}+1}{2005^{2016}+1}< 1\end{matrix}\right.\)

\(\Rightarrow\) Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:

\(\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2015}+1+2004}{2005^{2016}+1+2004}\)

\(=\dfrac{2005^{2015}+2005}{2005^{2016}+2005}=\dfrac{2005\left(2005^{2014}+1\right)}{2005\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)

\(\Rightarrow\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)

Vậy \(B< A\)

Hay \(A>B\)

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(A=\dfrac{2005^{2005}+1}{2005^{2006}+1}< 1\)

\(A< \dfrac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\Rightarrow A< \dfrac{2005^{2005}+2005}{2005^{2006}+2005}\Rightarrow A< \dfrac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\Rightarrow A< \dfrac{2005^{2004}+1}{2005^{2005}+1}=B\)

\(A< B\)

Ta có : A = \(\dfrac{2005^{2005}+1}{2005^{2006}+1}\)

\(2005\)A = \(\dfrac{\left(2005^{2005}+1\right).2005}{2005^{2006}+1}\)

\(2005\)\(A\)= \(\dfrac{2005^{2006}+2005}{2005^{2006}+1}\)

\(2005\)\(A\)= \(\dfrac{2005^{2006}+1+2004}{2005^{2006}+1}\)

\(2005A=\dfrac{2005^{2006}+1}{2005^{2006}+1}+\dfrac{2004}{2005^{2006}+1}\)

\(2005A=1+\dfrac{2004}{2005^{2006}+1}\)

Tương tự như vậy với \(B\) ta đc

\(2005B=1+\dfrac{2004}{2005^{2005}+1}\)

Vì \(2005^{2006}+1>2005^{2005}+1\)

\(=>\) \(1+\dfrac{2004}{2005^{2006}+1}\)\(< \)\(1+\dfrac{2004}{2005^{2005}+1}\)

\(=>\)\(2005A< 2005B\)

\(=>\)\(A< B\)

Vậy \(A< B\)

æ để bài này cho t nhé đợi t thương lượng với chủ thớt r` làm :V

Bài này t làm lần thứ n rồi. Thấy đề là ngán hết muốn làm luôn.

A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1 => \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\) = \(\frac{2005^{2005}+2005}{2005^{2006}+2005}\)= \(\frac{2005.\left(2005^{2004}+1\right)}{2005.\left(2005^{2005}+1\right)}\) = \(\frac{2005^{2004}+1}{2005^{2005}+1}\) = B => A<B.

Ta thấy:A=\(\frac{2005^{2005+1}}{2005^{2006}+1}\)<1
Ta có:A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\)<\(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\)=\(\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=b
Vậy A<B
Chắc chắn 100%

cái này trong đề cương

\(\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

\(\dfrac{2005.2006-1}{2004.2006}=1-\dfrac{1}{2005.2006}\)

\(Vì\dfrac{1}{2004.2005}>\dfrac{1}{2005.2006}\Rightarrow1-\dfrac{1}{2004.2005}< 1-\dfrac{1}{2005.2006}\Rightarrow\dfrac{2004.2005-1}{2004.2005}< \dfrac{2005.2006-1}{2004.2006}\)

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