Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)
\(x+A=\frac{47}{42}\)
ta thấy :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=\frac{1}{1}-\frac{1}{6}\)
\(A=\frac{5}{6}\)
vậy \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
hay \(x+\frac{5}{6}=\frac{47}{42}\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}\)
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}.\)
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
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
1/2+1/6+1/12+1/20+1/30+1/42
=1/1×2+1/2×3+1/3×4+1/4×5+1/5×6+1/6×7
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7.
Tích mk nha
=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{110}\)
=\(\frac{1}{1.2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+......+\frac{1}{10.11}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{10}-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
=\(\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+...+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{10\cdot11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
= \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ ...... + \(\frac{1}{7.8}\)
= \(1\)\(-\)\(\frac{1}{8}\)
= \(\frac{7}{8}\)
thiếu bước :v
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}\)
\(=\frac{7}{8}\)
1/
A= 1/15+1/35+1/63+1/99+ ... + 1/9999
A=1/3.5+1/5.7+1/7.9+ ... +1/99.101
2A=2/3.5+2/5.7+2/7.9+ ... +2/99.101
2A=1/3-1/5+1/5-1/7+1/7-1/9+ ... + 1/99-1/101
2A=1/3-1/101
A=49/303
Sai thì thôi nhé
A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
A=1-1/7
A=6/7
B = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{15}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{63}\)
B = \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)+\frac{1}{63}\)
B = \(1+\frac{1}{5}+\frac{3}{40}+\frac{1}{63}\)
B = \(1\frac{11}{40}+\frac{1}{63}\)
B = \(1\frac{733}{2520}\)
nguyentuantai làm hòa với Nguyễn Đình Dũng phải chăng mục đích là lấy **** ko
\(\Rightarrow A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow A=1-\frac{1}{7}\)
\(\Rightarrow A=\frac{6}{7}\)
nhớ tick nha!!!!!!!!!!!!!
A = 1/1*2 +1/2*3 +1/3*4 +1/4*5+ 1/5*6 +1/6*7
A = 1/1 - 1/2 +1/2 -1/3 +1/3 -1/4 +1/4 -1/5 +1/5 -1/6 +1/6 -1/7
A = 1 - 1/7
A= 6/7