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12 tháng 8 2018

a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)

      \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

       \(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

                  \(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

                   \(=3.\left(1-\frac{1}{100}\right)\)

                     \(=3.\frac{99}{100}=\frac{297}{100}\)

14 tháng 6 2015

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)

29 tháng 5 2018

\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

29 tháng 5 2018

Các bạn giải giúp mình nha😐

2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)

3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)

4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)

5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)

6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)

22 tháng 7 2016

1)C= 1/5+1/10+1/20+1/40+...+1/1280

\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)

Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)

\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)

\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)

2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10

\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=3\left(1-\frac{1}{10}\right)\)

\(=3\times\frac{9}{10}\)

\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé 

18 tháng 4 2020

Ghi đầy đủ nha

6 tháng 3 2022

bn có thể ghi rõ ràng đc ko?

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{11}{33}-\dfrac{35}{40}\)

`=`\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{1}{3}-\dfrac{7}{8}\)

`=`\(\dfrac{12}{24}-\dfrac{20}{24}+\dfrac{8}{24}-\dfrac{21}{24}\)

`= -21/24 = -7/8`

`b)`

\(\dfrac{2}{3}\cdot1\dfrac{3}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{2}{3}\cdot\dfrac{7}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{7}{6}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{5}{18}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(-\dfrac{1}{18}-\dfrac{1}{5}=-\dfrac{23}{90}\)

`c)`

\(\dfrac{1}{2}\cdot2-2\dfrac{5}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(1-\dfrac{19}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{12}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{3}{14}-\dfrac{10}{15}=-\dfrac{37}{42}\)

`d) `

\(\dfrac{1}{6}\cdot\dfrac{1}{11}+\dfrac{4}{11}\cdot\left(-\dfrac{1}{6}\right)+\dfrac{8}{11}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{6}{11}\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1}{11}-\dfrac{4}{11}+\dfrac{8}{11}+\dfrac{6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1-4+8+6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot1=\dfrac{1}{6}\)

`e)`

\(-17\cdot\left(-23\right)+\left(-53\right)\cdot17+17\cdot14+17\cdot\left(-24\right)\)

`= 17*(23-53+14-24)`

`= 17*(-40)`

`= -680`

`f)`

\(-19\cdot218+\left(-82\right)\cdot19-533\cdot19+\left(-19\right)\cdot167\)

`= 19*(-218-82-533-167)`

`= 19*(-1000)`

`= -19000`

`g)`

\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{11}{44}+\dfrac{9}{16}\)

`=`\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{31}{40}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{21}{40}+\dfrac{9}{16}=\dfrac{87}{80}\)

`h)`

\(\dfrac{4}{10}-1\dfrac{5}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{3}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{49}{15}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{287}{120}-\dfrac{1}{9}=-\dfrac{901}{360}\)

`i )`

\(3\cdot\dfrac{1}{5}-\dfrac{2}{8}-\dfrac{12}{36}+\dfrac{15}{9}\)

`=`\(\dfrac{3}{5}-\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{7}{20}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{1}{60}+\dfrac{15}{9}=-\dfrac{33}{20}\)

`k)`

\(\dfrac{6}{8}\cdot3\dfrac{1}{2}+4\dfrac{2}{3}-\dfrac{11}{55}+\dfrac{17}{51}\)

`=`\(\dfrac{3}{4}\cdot\dfrac{7}{2}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{21}{8}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{175}{24}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{851}{120}+\dfrac{17}{51}=\dfrac{297}{40}\)

`l )`

\(\dfrac{1}{3}\cdot3\dfrac{1}{2}-4\dfrac{2}{5}-\dfrac{26}{78}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{7}{2}-\dfrac{22}{5}-\dfrac{1}{3}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\left(\dfrac{7}{2}-1\right)-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{5}{2}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{5}{6}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(-\dfrac{107}{30}+\dfrac{17}{51}=-\dfrac{97}{30}\)

P/s: Bạn tách bài ra hỏi nhé! Và ghi đề rõ ràng chứ đừng ghi ntnay, nhiều bạn nhìn vào rất khó nhìn!

`# \text {KaizulvG}`

GH
26 tháng 6 2023

6 tháng 3 2023

`1/3 + 1/5 -1/4`

`= 5/15 + 3/15 -1/4`

`= 8/15 -1/4`

`=  17/60`

__

`7/8 -(1/4 + 2/5)`

`= 7/8-1/4 - 2/5`

`= 7/8 - 2/8 -2/5`

`= 5/8 -2/5`

`= 9/40`

__

`5/11-(3/5 -6/11)`

`= 5/11 -3/5 +6/11`

`= (5/11 + 6/11)-3/5`

`= 1 -3/5`

`= 5/5-3/5`

`=2/5`

__

`5/7 - ( 19/35 - 6/15)`

`= 5/7 - 19/35 + 6/15`

`= 25/35 -19/35 + 6/15`

`=6/35 +6/15`

`=4/7`

6 tháng 3 2023