\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A=\dfrac{1}{2}.\dfrac{100}{101}\)
\(A=\dfrac{50}{101}\)
Bạn xem Sách bài tập toán 6 tập 2 trang 24 bài 9.3b; 9.4; 9.5; 9.6 sẽ có
Nhớ tick cho mình nha
A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{99.101}\)
= \(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
=\(\dfrac{1}{2}.\dfrac{100}{101}\)
=\(\dfrac{50}{101}\)
Ta có: A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
⇒ 2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
2A=\(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{101-99}{99.101}\)
2A=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
2A=\(1-\dfrac{1}{101}\)
2A=\(\dfrac{100}{101}\)
⇒ A=\(\dfrac{100}{101}:2\)
⇒ A=\(\dfrac{50}{101}\)
