Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt biểu thức đó là A
=> 2A = \(\frac{2}{9.11}+\frac{2}{11.13}+...+\frac{2}{61.63}=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{61}-\frac{1}{63}\)
\(=\frac{1}{9}-\frac{1}{63}=\frac{2}{21}\)
=> A = \(\frac{2}{21}.\frac{1}{2}=\frac{1}{21}\)
Tính :
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\Rightarrow\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{5}{15}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)
A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\)
A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )
A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )
A= 5. (\(1-\frac{1}{100}\))
A= 5.\(\frac{99}{100}\)
A= \(\frac{99}{20}\)
B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)- \(\frac{1}{10}\)
= \(\frac{1}{2}\) - \(\frac{1}{10}\)
= \(\frac{2}{5}\)
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
= 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15 + 1/15 - 1/19
= 1/9 - 1/19
= 10/171
Tk mk nha
\(M=\frac{3}{2}.\left(\frac{2}{11.13}+\frac{2}{13.15}+......+\frac{2}{97.99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+.....+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{99}\right)=\frac{3}{2}.\frac{8}{99}=\frac{4}{33}\)
M= \(\frac{3}{11\cdot13}+\frac{3}{13\cdot15}+\frac{3}{15\cdot17}+...+\frac{3}{97\cdot99}\)
=\(\frac{3}{2}\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+...+\frac{2}{97\cdot99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\frac{8}{99}\)
= \(\frac{4}{33}\)
\(a.\frac{-5}{9}+\frac{5}{9}:A=\left(-\frac{5}{9}+\frac{5}{9}\right):A=0:A=0\)
\(b.\frac{7}{25}.\frac{11}{13}-\frac{7}{25}.\frac{2}{13}-\frac{8}{25}=\frac{7}{25}.\left(\frac{11}{13}-\frac{2}{13}\right)-\frac{8}{25}=\frac{7}{25}.\frac{9}{13}-\frac{8}{25}=\frac{63}{325}-\frac{8}{25}=\frac{-41}{325}\)
C.Thua
Đúng 100%
Đúng 100%
Đúng 100%
\(\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}+......+\frac{1}{97.99}\)
=\(\frac{1}{2}.\left(\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+......+\frac{2}{97.99}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+......+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{99}\right)=\frac{1}{2}.\left(\frac{11}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{10}{99}=\frac{5}{99}\)