\(B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+\dfrac{2}{6\cdot7\cdot8}\)
1)A=\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+.....+\dfrac{5}{99\cdot100}\)
C=\(1\cdot2\cdot3+2\cdot3\cdot4++3\cdot4\cdot5+4\cdot5\cdot6+5\cdot6\cdot7+6\cdot7\cdot8+7\cdot8\cdot9+8\cdot9\cdot10\)
D=\(1^2+2^2+3^2+...+99^2+100^2\)
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
\(\dfrac{1\cdot2\cdot3+2\cdot4+6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\)
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
Bạn ghi đề sai rồi, mình sửa lại đề ở phần (*) rồi nhé!
Ta có: \(\dfrac{1.2.3+2.4.6+4.8.12}{1.3.5+2.6.10+4.12.20}\) (*)
= \(\dfrac{1.2.3\left(1+2^3+4^3\right)}{1.3.5\left(1+2^3+4^3\right)}\) = \(\dfrac{1.2.3}{1.3.5}\) = \(\dfrac{2}{5}\)
\(\dfrac{1}{1\cdot2}\)+\(\dfrac{1}{2\cdot3}\)+\(\dfrac{1}{3\cdot4}\)+\(\dfrac{1}{4\cdot5}\)+\(\dfrac{1}{5\cdot6}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{5}-\dfrac{1}{6}\)
= 1 - \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\)
Bạn Huyền Nguyễn làm đúng òi. Bạn có thể xem cách giải cụ thể hơn ở sách bài tập nâng cao và một số chuyên đề toán 6 nhá!Trang 79. Ngày trước tớ cũng học mãi mới hiểu
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{2021\cdot2022\cdot2023}\)
\(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2021\cdot2022\cdot2023}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{2021\cdot2022\cdot2023}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2021\cdot2022}-\dfrac{1}{2022\cdot2023}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4090506}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2045252}{4090506}=\dfrac{1022626}{4090506}=\dfrac{511313}{2045253}\)
`1/(1.2.3) + 1/(2.3.4) + ... + 1/(2021 . 2022 .2023)`
`=> 2/(1.2.3) + 2/(2.3.4) + ... + 2/(2021 . 2022. 2023)`
`= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + ... + 1/(2021.2022) - 1/(2022 . 2023)`
`= 1/2 - 1/4090506`
`=4090506/8181012 - 2/8181012`
`= 4090504/8181012`
\(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\)
Có bao nhiêu \(A\left(A\inℕ^∗\right)\) mà \(1^9\cdot2^8\cdot3^7\cdot4^6\cdot5^5\cdot6^4\cdot7^3\cdot8^2\cdot9^1⋮A^2\)
\(\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\frac{1}{8\cdot7}-\frac{1}{7\cdot6}-\frac{1}{6\cdot5}-\frac{1}{5\cdot4}-\frac{1}{4\cdot3}-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
tính : \(A=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+\frac{2}{6\cdot7}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\left(\frac{7}{7}-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
Chúc bạn học tốt !!!
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{7}\right)\)
\(=2\times\frac{6}{7}\)
\(=\frac{12}{7}\)
Tk mk nha ~~
\(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+4\cdot8\cdot16+8\cdot16\cdot32}{1\cdot3\cdot4+2\cdot6\cdot8+4\cdot12\cdot16+8\cdot24\cdot32}\). dấu . là nhân .mọi ng cho m các phép tính cụ thể ạ.
\(\dfrac{1.2.4+2.4.8+4.8.16+8.16.32}{1.3.4+2.6.8+4.12.16+8.24.32}\)
\(=\dfrac{8.\left(1+8+4.16+16.32\right)}{12.\left(1+8+4.16+16.32\right)}\)
\(=\dfrac{8}{12}=\dfrac{2}{3}\)