Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài
Hoàng Đức Tùng

Thực hiện phép tính:
a) \(\left(\dfrac{1}{49}-\dfrac{1}{3^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{4^2}\right)...\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)\)

b) \(\dfrac{1}{999.1000}-\dfrac{1}{998.999}-\dfrac{1}{997.998}-...-\dfrac{1}{2.1}\)

c) \(\left(\dfrac{1}{2036}-\dfrac{1}{2006}\right)\left(\dfrac{1}{2031}-\dfrac{1}{2011}\right)\left(\dfrac{1}{2026}-\dfrac{1}{2016}\right)...\left(\dfrac{1}{2011}-\dfrac{1}{2031}\right)\)

d)\(\dfrac{\dfrac{4}{17}+\dfrac{4}{19}-\dfrac{4}{2111}}{\dfrac{5}{17}+\dfrac{5}{19}-\dfrac{5}{2111}}-\dfrac{\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}}{\dfrac{-5}{123}+\dfrac{5}{19}-\dfrac{5}{371}+1}\)

a: \(\left(\dfrac{1}{49}-\dfrac{1}{3^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{4^2}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)\)

\(=\left(\dfrac{1}{49}-\dfrac{1}{7^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{3^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{4^2}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)\)

\(=\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\left(\dfrac{1}{49}-\dfrac{1}{9}\right)\cdot\left(\dfrac{1}{49}-\dfrac{1}{16}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)\)

=0

b: \(\dfrac{1}{999\cdot1000}-\dfrac{1}{998\cdot999}-\dfrac{1}{997\cdot998}-...-\dfrac{1}{2\cdot1}\)

\(=\dfrac{1}{999\cdot1000}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{997\cdot998}+\dfrac{1}{998\cdot999}\right)\)

\(=\dfrac{1}{999}-\dfrac{1}{1000}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{998}-\dfrac{1}{999}\right)\)

\(=\dfrac{1}{999}-\dfrac{1}{1000}-1+\dfrac{1}{999}=\dfrac{2}{999}-\dfrac{1001}{1000}\)

\(=\dfrac{2000-1001\cdot999}{999\cdot1000}=\dfrac{-997999}{999000}\)

 

d: \(\dfrac{\dfrac{4}{17}+\dfrac{4}{19}-\dfrac{4}{2111}}{\dfrac{5}{17}+\dfrac{5}{19}-\dfrac{5}{2111}}-\dfrac{\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}}{-\dfrac{5}{123}+\dfrac{5}{19}-\dfrac{5}{371}+1}\)

\(=\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}-\dfrac{1}{2111}\right)}{5\left(\dfrac{1}{17}+\dfrac{1}{19}-\dfrac{1}{2111}\right)}-\dfrac{\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}}{-5\left(\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}\right)}\)

\(=\dfrac{4}{5}+\dfrac{1}{5}=1\)

Tui hổng có tên =33
2 tháng 9 lúc 15:17

\(\dfrac{\dfrac{4}{17}+\dfrac{4}{19}-\dfrac{4}{2111}}{\dfrac{5}{17}+\dfrac{5}{19}-\dfrac{5}{2111}}-\dfrac{\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}}{\dfrac{-5}{123}+\dfrac{5}{19}-\dfrac{5}{371}+1}\)
\(=\dfrac{4.\left(\dfrac{1}{17}+\dfrac{1}{19}-\dfrac{1}{2111}\right)}{5.\left(\dfrac{1}{17}+\dfrac{1}{19}-\dfrac{1}{2111}\right)}-\dfrac{\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}}{\dfrac{-5}{123}-\dfrac{-5}{19}+\dfrac{-5}{371}-\dfrac{-5}{5}}\)
\(=\dfrac{4}{5}-\dfrac{\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}}{\left(-5\right)\cdot\left(\dfrac{1}{123}-\dfrac{1}{19}+\dfrac{1}{371}-\dfrac{1}{5}\right)}\)
\(=\dfrac{4}{5}-\dfrac{1}{-5}\)
\(=\dfrac{4}{5}+\dfrac{1}{5}\)
\(=\dfrac{5}{5}=1\)


Các câu hỏi tương tự
Nguyễn Phương Chi
Xem chi tiết
Hang Nguyen
Xem chi tiết
quan nguyen hoang
Xem chi tiết
★彡✿ทợท彡★
Xem chi tiết
Xem chi tiết
Quỳnh nga
Xem chi tiết
fhdfhg
Xem chi tiết
Xem chi tiết
★彡✿ทợท彡★
Xem chi tiết
Trần Thị Thảo
Xem chi tiết