Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)

\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)

\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)

\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)

Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)

\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.

Vậy \(\dfrac{B}{A}\) là số nguyên.

Lời giải:

Ta có:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)

\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\)

\(\Leftrightarrow S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2016-2015}{2015.2016}\)

\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)

--------------------------

\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}\)

\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2014}{2015}\)

\(\Leftrightarrow S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2015-2014}{2014.2015}\)

\(\Leftrightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)

\(\Leftrightarrow S< 1-\frac{1}{2015}=\frac{2014}{2015}\)

Vậy ta có đpcm.

b\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4

Tương tự như vậy với câu a\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2

Lời giải:
$B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}$

$B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{8-7}{7.8}$

$B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}$

$B< 1-\frac{1}{8}$

Mà $1-\frac{1}{8}< 1$ nên $B< 1$ (đpcm)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{8^2}< 1\)

\(\dfrac{1}{2^2}< \dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

.......

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)

\(=>B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{8^2}< \dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.....+\dfrac{1}{7\cdot8}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+.....+\dfrac{1}{7}-\dfrac{1}{8}=1-\dfrac{1}{8}< 1\)

\(=>B< 1\)

Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...............

\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)

=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)

B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))

Vậy.....

Giải

Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)

Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)

\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

D< 1 - \(\dfrac{1}{20}\)

D< \(\dfrac{19}{20}\)<1

\(\Rightarrow\)D< 1

Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1

A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)

A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)

Ta có :

\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :

\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1

A<\(\dfrac{49}{200}< \dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{2}\)

có gì sai xin mấy bạn chỉ bảo!!!

Ta thấy A > 0 (1)

Vì \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2015.2016}\)

\(\Rightarrow A>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}< 1\)(2)

Từ (1)(2) => 0 < A < 1

Vậy A không phải là số tự nhiên

Giải:

Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}>0_{\left(1\right)}.\) (do A là phân số dương).

Ta lại có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}.\)

\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2016.2016}.\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}.\)

\(< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}.\)

\(< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2015}-\dfrac{1}{2015}\right)-\dfrac{1}{2016}.\)\(< 1+0+0+0+...+0-\dfrac{1}{2016}.\)

\(< 1-\dfrac{1}{2016}.\)

\(< \dfrac{2015}{2016}.\)

\(\Rightarrow A< 1_{\left(2\right)}.\) (do \(\dfrac{2015}{2016}< 1\)).

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\) \(\Rightarrow0< A< 1.\)

\(\Rightarrow A\) không phải là số tự nhiên.

Vậy ta thu được \(đpcm.\)

~ Học tốt!!! ~

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B< 1-\dfrac{1}{8}=\dfrac{7}{8}< 1\)

mink nhanh nhất đó bạn,

ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1\times2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\times3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\times4}\)

. . . . . . .

\(\dfrac{1}{8^2}< \dfrac{1}{7\times8}\)

_________________________________

\(\Rightarrow\)\(B< \)\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{7}-\dfrac{1}{8}\)

\(\Rightarrow B< 1-\dfrac{1}{8}\)

\(\Rightarrow B< 1\)

\(\Rightarrowđpcm\)