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4 tháng 7 2017

\(-1-\dfrac{1}{3}-\dfrac{1}{6}-...-\dfrac{1}{1225}\)

\(=\dfrac{-1}{2}\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2450}\right)\)

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

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

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

\(=\dfrac{-1}{2}.\dfrac{49}{50}=\dfrac{-49}{100}\)

Vậy...

12 tháng 12 2023

Google

 

12 tháng 12 2023

A =  -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)

    = -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))

Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)

Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))

               = 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))

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

               = 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))

               = 2×\(\dfrac{24}{50}\)

                   =  \(\dfrac{24}{25}\)

      Thay B vào A ta có :

   A = -1-\(\dfrac{24}{25}\)

 => A = \(\dfrac{-49}{25}\)

 Cho mik một tick nhé thankss

14 tháng 9 2023

\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)

\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)

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

\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)

\(\dfrac{1}{2}B=\dfrac{-49}{50}\)

\(B=\dfrac{-49}{25}\)

14 tháng 9 2023

\(B=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)

\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)

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

=-2*49/50

=-49/25

15 tháng 11 2025

a: \(A=\left(\frac{136}{15}-\frac{28}{5}+\frac{62}{10}\right)\cdot\frac{21}{24}\)

\(=\left(\frac{272}{30}-\frac{168}{30}+\frac{186}{30}\right)\cdot\frac78\)

\(=\frac{290}{30}\cdot\frac78=\frac{29}{3}\cdot\frac78=\frac{203}{24}\)

b: \(B=\frac56+6\frac56\left(11\frac{5}{20}-9\frac14\right):8\frac13\)

\(=\frac56+\frac{41}{6}\cdot\left(11+\frac14-9-\frac14\right):\frac{25}{3}=\frac56+\frac{41}{6}\cdot2\cdot\frac{3}{25}\)

\(=\frac56+\frac{41}{25}=\frac{125}{150}+\frac{246}{150}=\frac{371}{150}\)

c: \(C=1+3+6+\cdots+1225\)

\(=\frac12\left(2+6+12+\cdots+2450\right)=\frac12\cdot\left(1\cdot2+2\cdot3+\cdots+49\cdot50\right)\)

\(=\frac12\cdot\left\lbrack1\left(1+1\right)+2\left(2+1\right)+\cdots+49\left(49+1\right)\right\rbrack=\frac12\cdot\left\lbrack\left(1+2+\cdots+49\right)+\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot\frac{50}{2}+\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot25+49\cdot50\cdot\frac{99}{6}\right\rbrack=\frac12\cdot\left\lbrack49\cdot25+49\cdot25\cdot33\right\rbrack=\frac12\cdot49\cdot25\cdot\left(33+1\right)\)

\(=49\cdot25\cdot\frac{34}{2}=49\cdot25\cdot17=20825\)

2 tháng 4

21 tháng 11 2018

a) \(A=\dfrac{5^2}{11.16}+\dfrac{5^2}{16.21}+\dfrac{5^2}{21.26}+...+\dfrac{5^2}{56.61}\)

\(A=5^2.\left(\dfrac{1}{11.16}+\dfrac{1}{16.21}+\dfrac{1}{21.26}+...+\dfrac{1}{56.61}\right)\)

\(A=\left(5^2:5\right).\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+...+\dfrac{5}{56.61}\right)\)

\(A=5.\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+...+\dfrac{1}{56}-\dfrac{1}{61}\right)\)

\(A=5.\left(\dfrac{1}{11}-\dfrac{1}{61}\right)\)

\(A=5.\dfrac{50}{671}\)

\(Á=\dfrac{250}{671}\)

22 tháng 11 2022

b: \(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)

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

\(=-2\cdot\dfrac{49}{50}=-\dfrac{49}{25}\)

14 tháng 5 2022

cíu mik cần gấp

15 tháng 5 2022

thế này mà là lớp 1 ! ảo thật đấy 

25 tháng 6 2023

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

6 tháng 4 2022

\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)

6 tháng 4 2022

=\(\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)