Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)
\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)
Còn lại em nhân giống ở trên nhé
Đặt A = 1/15 + 1/35 + ... + 1/3135
A = 1/3.5 + 1/5.7 + ... + 1/55.57
2A = 2/3.5 + 2/5.7 + ... + 2/55.57
2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57
2A = 1/3 - 1/57 = 6/19
A = 3/19
\(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot1\frac{1}{24}\cdot...\cdot1\frac{1}{99}\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot\frac{25}{24}\cdot...\cdot\frac{100}{99}\)
\(=\frac{2.2\cdot3.3\cdot4.4\cdot5.5\cdot...\cdot10.10}{1.3\cdot2.4\cdot3.5\cdot4.6\cdot...\cdot9.11}\)
\(=\frac{2.10}{1.11}=\frac{20}{11}\)
"." = nhân
a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)
=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)
=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)
=> \(x-\frac{11}{5}=2\)
=> \(x=2+\frac{11}{5}\)
=> \(x=\frac{21}{5}\)
( 4.2/5 +3/5) + (2.3/4 +1/4) +(5.2/3 +1/3) =( 4 + 2/5 + 3/5) +( 2 +3/4 + 1/4 ) + (5 + 2/3 + 1/3) =( 4+1 ) + (2 +1) + ( 5 + 1) = 14
b. \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2003}\)= \(\frac{1}{2013}\)
c. \(5.\left(\frac{1}{14}+\frac{1}{84}+\frac{1}{204}+\frac{1}{374}\right)\)= 5. \(\frac{1}{11}\)= \(\frac{5}{11}\)
Mình biết 2 câu này thôi, thông cảm nhá...!!!
d) Ta có: 100-(1+1/2+1/3+1/4+...+1/100)
=1x100-(1+1/2+1/3+1/4+...+1/100)
=(1-1)+(1-1/2)+(1-1/3)+(1-1/4)+....+(1-1/100)
=1/2+2/3+3/4+...+99/100
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)
\(\Leftrightarrow x=122\)
