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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
\(\left(y-\frac{1}{2}\right):\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
=> \(\left(y-\frac{1}{2}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
=> \(\left(y-\frac{1}{2}\right):\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
=> \(\left(y-\frac{1}{2}\right):\left(1-\frac{1}{10}\right)=\frac{1}{3}\)
=> \(\left(y-\frac{1}{2}\right):\frac{9}{10}=\frac{1}{3}\)
=> \(y-\frac{1}{2}=\frac{3}{10}\)
=> \(y=\frac{13}{10}\)
Study well ! >_<
1/6 + 1/12 + 1/20 + 1/30 + 1/42 + ... + 1/90 + 1/110 = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + ... + 1/9.10 + 1/10.11 = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/9 - 1/10 + 1/10 - 1/11 = 1/2 - 1/11 = 9/22
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
Chỗ cuối mk nhầm, sửa lại nha :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{100}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{50}{100}-\frac{10}{100}+\frac{1}{100}\)
\(=\frac{41}{100}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{100}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{50}{100}-\frac{10}{100}-\frac{1}{100}\)
\(=\frac{39}{100}\)
Đúng thì k nha bn !!!!
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
khong the co n vi 40 khong bang so tu nhien lien tiep nao ca
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