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

khong the co n vi 40 khong bang so tu nhien lien tiep nao ca

n=1560

tic.k nha

bài này ra 1560 , cô mk giải đó , viết cách làm ra ngại lém

Ta có: 1/2=1/1*2                                                                m*m+1=n

Nên 1/1*2 +  1/2*3  +  1/3*4  +...  +1/m*m+1

(1-1/2) + (1/2-1/3) + ... (1/m-1/m+1)

Triệt tiêu đi còn1- 1/m+1 =39/40

Suy ra 1/m+1 = 1/40

Vậy m=39

n = 39* (39+1) = 39*40= 1560

\(\left(y-\frac{1}{2}\right):\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)

=> \(\left(y-\frac{1}{2}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)

=> \(\left(y-\frac{1}{2}\right):\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)

=> \(\left(y-\frac{1}{2}\right):\left(1-\frac{1}{10}\right)=\frac{1}{3}\)

=> \(\left(y-\frac{1}{2}\right):\frac{9}{10}=\frac{1}{3}\)

=> \(y-\frac{1}{2}=\frac{3}{10}\)

=> \(y=\frac{13}{10}\)

Study well ! >_<

giai câu a

a)   ta có   (\(\frac{2}{11.13}\)+\(\frac{2}{13.15}\)+.....+\(\frac{2}{19.21}\))*462 - x =19

                (\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{19}\)-\(\frac{1}{21}\)) * 462 -x=19

                (\(\frac{1}{11}\)-\(\frac{1}{21}\))*462-x=19

còn lại ban tu tinh nhu bai tim x

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

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

\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)

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

                                      \(=\frac{1}{1}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

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

\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)