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a)Ta có:  \(\frac{1}{3.4}=\frac{1}{12}\)

và \(\frac{1}{3}-\frac{1}{4}=\frac{4}{12}-\frac{3}{12}=\frac{1}{12}\)

\(\Rightarrow\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

b)Ta có : \(\frac{1}{5.6}=\frac{1}{30}\)

và \(\frac{1}{5}-\frac{1}{6}=\frac{6}{30}-\frac{5}{30}=\frac{1}{30}\)

\(\Rightarrow\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)

hok tốt!!

1 tháng 7 2015

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\)

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

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Do \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{100}\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>25\cdot\frac{1}{80}+25\cdot\frac{1}{100}=\frac{7}{12}\)

và \(A

1 tháng 7 2015

olm lag kinh đang làm lag thoát ra mất tiêu

-------đề đúng------------

19 tháng 9 2023

\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)+\left(\dfrac{5}{6}+\dfrac{19}{20}+...+\dfrac{2549}{2550}\right)\)

\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{50\cdot51}\right)+\left(1-\dfrac{1}{2\cdot3}\right)+\left(1-\dfrac{1}{3\cdot4}\right)+...+\left(1-\dfrac{1}{50\cdot51}\right)\)

\(B=\left(1+1+...+1\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\)

\(B=1\cdot49=49\) (vì có (50 - 2) : 1 + 1 = 49 số hạng 1)

2 tháng 4 2020

\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)

\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=\frac{1}{2}-\frac{1}{6}\)

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

19 tháng 3 2022

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{21-2}{42}=\dfrac{19}{42}\)

AH
Akai Haruma
Giáo viên
19 tháng 3 2022

Lời giải:
Gọi biểu thức số 1 là A và số 2 là B

\(A=\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

B tương tự A:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)

\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)

21 tháng 6 2018

\(M=\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\)\(-\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)

\(M=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}\right)\)\(-\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)

\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}\)

\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{2}{6}-\frac{2}{8}-\frac{2}{10}\)

\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\)

\(M=1-\frac{1}{2}-\frac{2}{4}\)

\(M=1-\frac{1}{2}-\frac{1}{2}\)

\(M=0\)

              HOK TỐT

30 tháng 5 2023

\(\left(x+\dfrac{1}{2\times3}\right)+\left(x+\dfrac{1}{3\times4}\right)+\left(x+\dfrac{1}{4\times5}\right)+\left(x+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)

\(x+\dfrac{1}{2\times3}+x+\dfrac{1}{3\times4}+x+\dfrac{1}{4\times5}+x+\dfrac{1}{5\times6}=\dfrac{25}{3}\)

\(x\times4+\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)

\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)

\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)

\(x\times4+\dfrac{4}{12}=\dfrac{25}{3}\)

\(x\times4=\dfrac{25}{3}-\dfrac{4}{12}\)

\(x\times4=\dfrac{25}{3}-\dfrac{1}{3}\)

\(x\times4=\dfrac{24}{3}\)

\(x\times4=8\)

\(x=8\div4\)

\(x=2\)

:))

 

 

=1+1/3-1/4+1/4-1/5+1/5-1/6

=4/3-1/6=7/6

16 tháng 4 2022

Đợi nhé

10 tháng 6 2015

1> a) \(\frac{5}{7}x4:\frac{5}{9}=\frac{5}{7}:\frac{5}{9}x4=\frac{5}{7}x\frac{9}{5}x4=\frac{9}{7}x4=\frac{9x4}{7}=\frac{36}{7}\)

\(b,8x\frac{2}{3}:\frac{1}{2}=8x\frac{2}{3}x\frac{2}{1}=8x2x\frac{2}{3}=16x\frac{2}{3}=\frac{32}{3}\)

\(c,6:\frac{3}{5}-\frac{7}{6}x\frac{6}{7}=6x\frac{5}{3}-1=10-1=9\)

\(\frac{21}{5}x\frac{10}{11}+\frac{57}{11}=\frac{42}{11}+\frac{57}{11}=\frac{99}{11}=9\)

2) a) \(\frac{35}{9}:x=\frac{35}{6}\)

\(x=\frac{35}{9}:\frac{35}{6}\)

\(x=\frac{35}{9}x\frac{6}{35}\)

\(x=\frac{2}{3}\)

b) \(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\right)x10-X=0\)

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{5}-\frac{1}{6}\right)x10-X=0\)

\(\left(\frac{1}{1}-\frac{1}{6}\right)x10-X=10\)

\(\frac{5}{6}x10-X=0\)

\(X=\frac{5}{6}x10=\frac{25}{3}\)

Đúng nha !!!!

10 tháng 6 2015

1/a/\(\frac{5}{7}\cdot4:\frac{5}{9}=\frac{20}{7}:\frac{5}{9}=\frac{20}{7}\cdot\frac{9}{5}=\frac{36}{7}\)

b/\(8\cdot\frac{2}{3}:\frac{1}{2}=\frac{16}{3}:\frac{1}{2}=\frac{16}{3}\cdot\frac{2}{1}=\frac{32}{3}\)

c/\(6:\frac{3}{5}-\frac{7}{6}\cdot\frac{6}{7}=6\cdot\frac{5}{3}-1=10-1=9\)

2/a/\(\frac{35}{9}:x=\frac{35}{6}\)

\(x=\frac{35}{9}:\frac{35}{6}=\frac{35}{9}\cdot\frac{6}{35}\)

\(x=\frac{2}{3}\)

 

b/\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\cdot10-x=0\)

\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\cdot10-x=0\)

\(\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{2}{30}\right)\cdot10-x=0\)

\(\frac{47}{60}\cdot10-x=0\)

\(\frac{47}{6}-x=0\)

\(x=\frac{47}{6}-0\)

\(x=\frac{47}{6}\)