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19 tháng 5 2020

A=\(\frac{3}{7}\)

CÒN CÁCH LÀM ĐANG CHƯA BIẾT

19 tháng 5 2020

Trả lời:

\(A=\frac{3}{7}\)

Hmm chứ ko phải là cứ cộng hết vào là đc ạ hay phải tính nhanh?

:p

11 tháng 7 2018

Tính nhanh : 

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(A=2\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+\frac{1}{10\cdot12}+\frac{1}{12\cdot14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)

\(A=2\cdot\frac{3}{7}\)

\(A=\frac{6}{7}\)

11 tháng 7 2018

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)

\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)

\(A=\frac{1}{2}-\frac{1}{14}\)

\(A=\frac{3}{7}\)

_Chúc bạn học tốt_

8 tháng 7 2018

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

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

\(=\frac{1}{2}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}=\frac{3}{7}\)

8 tháng 7 2018

Đặt \(C=\frac{1}{2}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{84}\)

\(\Rightarrow\frac{C}{2}=1+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{7}\)

\(\Rightarrow C=\left(1+\frac{1}{2}-\frac{1}{7}\right).2\)

9 tháng 7 2015

Ax1/2=1/2+1/6+1/12+1/20+1/30+1/42

Ax1/2=1/1.2+1/2.3+1/3.4+...+1/6.7

Ax1/2=1-1/2+1/2-1/3+1/3-1/4+....+1/6-1/7

Ax1/2=1-1/7=6/7

=>A=6/7x2=12/7

2 tháng 4 2020

A= \(\frac{1}{4}\)\(\frac{1}{12}\)\(\frac{1}{24}\)\(\frac{1}{40}\)\(\frac{1}{60}\)\(\frac{1}{84}\)
2A=\(\frac{1}{2}\)\(\frac{1}{6}\)\(\frac{1}{12}\)+\(\frac{1}{20}\)\(\frac{1}{30}\)\(\frac{1}{42}\)
2A=\(\frac{1}{1.2}\)\(\frac{1}{2.3}\)\(\frac{1}{3.4}\)\(\frac{1}{4.5}\)\(\frac{1}{5.6}\)\(\frac{1}{6.7}\)
2A=\(1-\frac{1}{2}\)\(+\frac{1}{2}\)\(-\frac{1}{3}\)\(+...+\)\(\frac{1}{6}\)\(-\frac{1}{7}\)
2A=\(1-\frac{1}{7}\)
2A=\(\frac{6}{7}\)
A=\(\frac{6}{7}:2\)
\(\Rightarrow A=\frac{3}{7}\)
Vậy \(A=\frac{3}{7}\)
Chúc bạn học tốt nhé!

17 tháng 8 2016

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

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

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

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

9 tháng 2 2018

\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)

\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

\(B=\frac{3}{4}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)

\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)

\(A=\frac{2}{3}-\frac{1}{192}\)

\(A=\frac{127}{192}\)

\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

      \(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)

      \(C=\frac{1990.997}{1994.995}\)

      \(C=\frac{995.2+997}{997.2+995}=1\)

9 tháng 2 2018

\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)

\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)

A = 1/1x2 +1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + .... + 1/y x n = 39/40

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/y - 1/n = 39/40

A = 1 - 1/n = 39/40

A = 1 - 39/40 = 1/n 

A = 1/40 = 1/n

=> n = 40

Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n}\) = \(\frac{39}{40}\)

= \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{ax\left(a+1\right)}=\frac{39}{40}\)  ( có : n = a x ( a+1) )

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}=\frac{39}{40}\)

=\(\frac{1}{1}-\frac{1}{a+1}=\frac{39}{40}\)

( ta triệt tiêu tất cả các phân số ở giữa )   VD: trừ 1/2 rồi lại cộng 1/2 thì còn lại 0 

\(\frac{1}{a+1}=\frac{1}{1}-\frac{39}{40}\)

\(\frac{1}{a+1}=\frac{1}{40}\)

a+1 = 40

a  = 40 - 1

a = 39

vì a x (a+1) = n

nên 39 x 40 = n

      n = 1560 

                      \(ĐS:1560\)

CHÚC BẠN HỌC GIỎI

19 tháng 7 2018

a)     5/30+15/90+25/150+35/210+45/270

       =1/6+1/6+1/6+1/6+1/6

       =1/6 x 5

       =5/6

b)     1/2+1/6+1/12+1/20+....+1/56

        =1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8

        =1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8

        =1/1-1/8

         =7/8

c)     mình chịu

19 tháng 7 2018

thank you bn nhìu nha