\(A=\dfrac{2017-2n}{8n-4}\)

\(A=-\dfrac{\left(2n-1\right)-2016}{4\left(2n-1\right)}\)

\(A=-\dfrac{1}{4}+\dfrac{2016}{4\left(2n-1\right)}\)

\(A=-\dfrac{1}{4}+\dfrac{504}{2n-1}\)

Để A có GTLN thì \(\dfrac{504}{2n-1}\) lớn nhất `=>2n-1` nhỏ nhất

`=>2n-1` là số nguyên dương nhỏ nhất

`=>2n-1=1` `=>n=1`

\(\Rightarrow A=-\dfrac{1}{4}+504=\dfrac{2015}{4}\)

Vậy \(Max_A=\dfrac{2015}{4}\) khi `n=1`

Rút gọn A

\(A=\dfrac{2017-2n}{8n-4}=\dfrac{2013+2\left(2-n\right)}{4\left(n-2\right)},n\ne2\)

\(A=\dfrac{2013+2\left(2-n\right)}{4\left(n-2\right)}=-\dfrac{2\left(n-2\right)}{4\left(n-2\right)}+\dfrac{2013}{4\left(n-2\right)}=-\dfrac{1}{2}+\dfrac{2013}{4\left(n-2\right)}\)

Để A có giá trị lớn nhất thì \(\dfrac{2013}{4\left(n-2\right)}\) lớn nhất, khi đó 4(n - 2) nhỏ nhất lớn hơn 0 hay n - 2 nhỏ nhất lớn hơn 0. Vì n nguyên nên n - 2 = 1 nhỏ nhất , suy ra n =3.

Vậy với n = 3 thì A đạt giá trị lớn nhất và A = -1/2 + 2013/4

chưa có ai giúp em sao?

A = \(\dfrac{1}{99}\) - \(\dfrac{1}{99\times97}\) - \(\dfrac{1}{97\times95}\) - \(\dfrac{1}{95\times93}\)-....-\(\dfrac{1}{5\times3}\)-\(\dfrac{1}{3\times1}\)

A = \(\dfrac{1}{99}\) - \(\dfrac{1}{2}\) .( \(\dfrac{2}{99\times97}\) + \(\dfrac{2}{97\times95}\)+ \(\dfrac{2}{95\times93}\)+......+\(\dfrac{2}{5\times3}\)+ \(\dfrac{2}{3\times1}\))

A = \(\dfrac{1}{99}\) - \(\dfrac{1}{2}\). (\(\dfrac{2}{1.3}\)+ \(\dfrac{2}{3.5}\)+ \(\dfrac{2}{5.7}\)+........+\(\dfrac{2}{93.95}\)+ \(\dfrac{2}{95.97}\)+\(\dfrac{2}{97.99}\))

A = \(\dfrac{1}{99}\) - \(\dfrac{1}{2}\). ( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+ .....+\(\dfrac{1}{93}\)-\(\dfrac{1}{95}\)+ \(\dfrac{1}{95}\)- \(\dfrac{1}{97}\)+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\))

A = \(\dfrac{1}{99}\)- \(\dfrac{1}{2}\) . (1 - \(\dfrac{1}{99}\))

A = \(\dfrac{1}{99}\) - \(\dfrac{1}{2}\). \(\dfrac{98}{99}\)

A = \(\dfrac{1}{99}\)- \(\dfrac{49}{99}\)

A = - \(\dfrac{48}{99}\)

A =  \(\dfrac{1}{99}\) - \(\dfrac{1}{99.98}\) - \(\dfrac{1}{98.97}\)- \(\dfrac{1}{97.96}\) -......-\(\dfrac{1}{3.2}\)- \(\dfrac{1}{2.1}\)

A = \(\dfrac{1}{99}\) - ( \(\dfrac{1}{99.98}\)+ \(\dfrac{1}{98.97}\)+\(\dfrac{1}{97.96}\)+....+\(\dfrac{1}{3.2}\)+\(\dfrac{1}{2.1}\))

A =  \(\dfrac{1}{99}\) - ( \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ ......+ \(\dfrac{1}{96.97}\)+\(\dfrac{1}{97.98}\)+ \(\dfrac{1}{98.99}\))

A = \(\dfrac{1}{99}\) - ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ......+ \(\dfrac{1}{96}\)- \(\dfrac{1}{97}\)+ \(\dfrac{1}{97}\)- \(\dfrac{1}{98}\)+ \(\dfrac{1}{98}\) - \(\dfrac{1}{99}\))

A = \(\dfrac{1}{99}\)- ( \(\dfrac{1}{1}\)- \(\dfrac{1}{99}\))

A = \(\dfrac{1}{99}\) - \(\dfrac{98}{99}\)

A = - \(\dfrac{97}{99}\)

\(D=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1255}\right)\)

\(\dfrac{1}{2}D=-\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{2.6}+\dfrac{1}{2.10}+\dfrac{1}{2.15}+...+\dfrac{1}{2.1255}\right)\)

\(\dfrac{1}{2}D=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\right)\)

\(\dfrac{1}{2}D=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\dfrac{1}{2}D=-\left(1-\dfrac{1}{50}\right)=-\dfrac{49}{50}\)

\(D=-\dfrac{49}{50}:\dfrac{1}{2}=-\dfrac{49}{25}\)

Bạn tham khảo nhé.

\(C=-\dfrac{3}{11.14}-\dfrac{3}{14.17}-\dfrac{3}{17.20}-...-\dfrac{3}{98.101}\)

\(\Rightarrow C=-\left(\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+...+\dfrac{3}{98.101}\right)\)

\(\Rightarrow C=-\left(\dfrac{14-11}{11.14}+\dfrac{17-14}{14.17}+...+\dfrac{101-98}{98.101}\right)\)

\(\Rightarrow C=-\left(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{98}-\dfrac{1}{101}\right)\)

\(\Rightarrow C=-\left(\dfrac{1}{11}-\dfrac{1}{101}\right)=\dfrac{-90}{1111}\)

Cảm ơn bạn! Đúng lúc bạn đang trả lời thì mình lại biết cách làm rồi. Cũng giống như cách làm của bạn.

  ( 8- \(\dfrac{9}{4}\)+ \(\dfrac{2}{7}\)) -(-6 - \(\dfrac{3}{7}\)+ \(\dfrac{5}{4}\)) -( 3 + \(\dfrac{2}{4}\) - \(\dfrac{9}{7}\))

= 8 - \(\dfrac{9}{4}\)+ \(\dfrac{2}{7}\) + 6 + \(\dfrac{3}{7}\)- \(\dfrac{5}{4}\)- 3 - \(\dfrac{2}{4}\)+ \(\dfrac{9}{7}\)

= (8 + 6 -3) - ( \(\dfrac{9}{4}\)+ \(\dfrac{5}{4}\)+ \(\dfrac{2}{4}\)) + ( \(\dfrac{2}{7}\)+ \(\dfrac{3}{7}\)+ \(\dfrac{9}{7}\))

= 11 - \(\dfrac{16}{4}\)+ \(\dfrac{14}{7}\)

= 11 - 4 + 2

= 9

   4/3 - [(-11/6) - (2/9 + 5/3)]

= 4/3 - [(-11/6) - (2/9 + 15/9)]

= 4/3 - [(-11/6) - 17/9]

= 4/3 - [ -33/18 - 34/18]

= 4/3 - [ -67/18 ]

= 24/18 + 67/18

= 91/18

= 

-37/18

   \(\dfrac{x-3}{x-2}\) = \(\dfrac{4}{7}\)

⇒ (x - 3) x 7 = (x - 2) x 4

⇒     7x - 21 = 4x - 8

⇒     7x - 4x = 21 - 8

⇒            3x = 13

⇒              x = \(\dfrac{13}{3}\)

x = 13/3

xy-5y=-3

=> y(x-5)=-3

Mà x , y ∈ Z

=> y ; x-5 là các cặp ước của -3Ta có bảng :

C1: \(\dfrac{-4}{15}=\dfrac{-1-3}{3\times5}=\dfrac{-1}{15}+\dfrac{-1}{5}\)

C2: \(\dfrac{-4}{15}=\dfrac{-8}{30}=\dfrac{-3-5}{2\times3\times5}=\dfrac{-1}{10}+\dfrac{-1}{6}\)

C3: \(\dfrac{-4}{15}=\dfrac{-12}{45}=\dfrac{-3-9}{2\times3\times5\times3}=\dfrac{-1}{30}+\dfrac{-1}{10}\)