a.

$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$

$=1-\frac{1}{1000}=\frac{999}{1000}$

b.

$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$

$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$

$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$

$=1-\frac{1}{500}=\frac{499}{500}$

$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$

c.

$2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.100}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{998.999}-\frac{1}{999.1000}$

$=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{499499}{999000}$

$\Rightarrow C=\frac{499499}{999000}:2=\frac{499499}{1998000}$

2, a-b=ab => a=ab+b => a=b(a+1)

thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1

Theo bài ra ta có: a:b=a-b

=> a+1=a-b

=>-b=1

=> b=-1

Thay b=-1 vào a-b=ab ta có : a-(-1)=-a

=> a +1=-a

=>a=-1/2

Vậy a=-1/2. b=-1

Ta biến đổi \(A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{2016-2015}{2016.2015}+\dfrac{2018-2017}{2017.2018}\) 

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1009}\right)\)

\(A=\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2017}+\dfrac{1}{2018}\)

Lại có \(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\)

\(B=\dfrac{1}{3028}.\left(\dfrac{3028}{1010.2018}+\dfrac{3028}{1011.2017}+...+\dfrac{3028}{2018.1010}\right)\)

\(B=\dfrac{1}{3028}\left(\dfrac{1}{1010}+\dfrac{1}{2018}+\dfrac{1}{1011}+\dfrac{1}{2017}+...+\dfrac{1}{2018}+\dfrac{1}{1010}\right)\)

\(B=\dfrac{1}{3028}.2\left(\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2018}\right)\)

\(B=\dfrac{1}{3028}.2A\) \(\Rightarrow\dfrac{A}{B}=1514\inℤ\). Ta có đpcm

a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)

b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)

Lời giải:

Ta có:

\(\frac{1}{1.2^2}=\frac{1}{2^2}\)

\(2.3^2>3^2\Rightarrow \frac{1}{2.3^2}< \frac{1}{3^2}\)

\(3.4^2> 4^2\Rightarrow \frac{1}{3.4^2}< \frac{1}{4^2}\)

...........

\(49.50^2> 50^2\Rightarrow \frac{1}{49.50^2}< \frac{1}{50^2}\)

Cộng theo từng vế các BĐT:

\(\Rightarrow \frac{1}{1.2^2}+\frac{1}{2.3^2}+\frac{1}{3.4^2}+....+\frac{1}{49.50^2}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}\)

\(\Leftrightarrow A< B\)

Vậy ta có đpcm.

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)