2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.

Xét:

`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`

`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`

`=>A+B>2`

Mà `1 2013/2014<2`

`=>A+B>1 2013/2014`

Ai trả lời đi please

A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B

\(\Rightarrow\) \(\dfrac{A}{B}\)=2015

Câu 2:

\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\)

\(=2014\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\)

\(=2014\left(1+\dfrac{1}{2\left(2+1\right)}.2+\dfrac{1}{3\left(3+1\right)}.2+...+\dfrac{1}{2013\left(2013+1\right)}.2\right)\)

\(=2014\left(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2013.2014}\right)\)

\(=4028\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\right)\)

Bạn tự tính nốt nhé

1)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\left(1\right)\)\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\\ =\dfrac{1}{1}-\dfrac{1}{2012}< 1\left(2\right)\)

Từ (1) và (2) ta có: A < 1

2)

\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\\ =2014\cdot\left(\dfrac{1}{1}+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\\ =2014\cdot\left(\dfrac{1}{\left(1\cdot2\right):2}+\dfrac{1}{\left(2\cdot3\right):2}+\dfrac{1}{\left(3\cdot4\right):2}+...+\dfrac{1}{\left(2013\cdot2014\right):2}\right)\\ =2014\cdot\left(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{2013\cdot2014}\right)\\ =2014\cdot2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2013\cdot2014}\right)\\ =4028\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\right)\\ =4028\cdot\left(1-\dfrac{1}{2014}\right)\\ =4028\cdot\dfrac{2013}{2014}\\ =4026\)

3)

Để A là số nguyên thì \(6n+42⋮6n\Rightarrow42⋮6n\Rightarrow6n\inƯ\left(42\right)\)

\(Ư\left(42\right)=\left\{1;2;3;6;7;14;21;42\right\}\)

Vì n là số tự nhiên nên n = 1 hoặc n = 7

4)

\(A=\dfrac{17^{18}+1}{17^{19}+1}< \dfrac{17^{18}+1+16}{17^{19}+1+16}=\dfrac{17^{18}+17}{17^{19}+17}=\dfrac{17\cdot\left(17^{17}+1\right)}{17\cdot\left(17^{18}+1\right)}=\dfrac{17^{17}+1}{17^{18}+1}=B\)

Vậy A<B

\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\)

\(\Rightarrow5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\)

\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\right)\)

\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)

\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}< \dfrac{1}{4}\)

\(\Rightarrowđpcm\)

\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)

\(\Rightarrow5A=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)

\(\Rightarrow5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\)

\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)

\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)

\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}\)

\(\Rightarrow A< \dfrac{1}{4}\)

\(A=\dfrac{2}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\\ =\dfrac{2}{4}.\left(\dfrac{1}{3}-\dfrac{1}{n+4}\right)\\ =\dfrac{1}{2}.\dfrac{n+1}{3\left(n+4\right)}=\dfrac{n+1}{6\left(n+4\right)}\\ =\dfrac{n+4-3}{6\left(n+4\right)}=\dfrac{1}{6}-\dfrac{1}{2\left(n+4\right)}< \dfrac{1}{6}.\)

Giải:

A=2/3.7+2/7.11+2/11.15+...+2/n.(n+4)

A=1/2.(4/3.7+4/7.11+4/11.15+...+4/n.(n+4)

A=1/2.(1/3-1/7+1/7-1/11+1/11-1/15+...+1/n-1/n+4)

A=1/2.(1/3-1/n+4)

A=1/6-1/2.(n+4)

⇒A>1/6

Chúc bạn học tốt!