Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\frac{3.8.15....9999}{4.9.16....10000}=?\)
Ta có:
B = \(\frac{-5}{3}+\frac{-5}{15}+\frac{-5}{35}+...+\frac{-5}{2499}.\)
= \(-5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\right)\)
= \(-5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
= \(-5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right):2\)
= \(\frac{-5}{2}\left(1-\frac{1}{51}\right)\)
= \(\frac{-5}{2}.\frac{50}{51}\) = -6375
Đặt A = (1/2)(3/4)(5/6) ... (9999/10000) (A > 0)
.Và B = (2/3)(4/5)(6/7) ... (10000/10001) (B > 0)
Ta có A.B = (1/2)(2/3)(3/4) ... (10000/10001) = 1/10001 (1)
Mặt khác :
1/2 < 2/3
3/4 < 4/5
................
................
9999/10000 < 10000/10001
Nhân tất cả vế theo vế ---> A < B ---> A² < A.B (2)
(1),(2) ---> A² < 1/10001 ---> A < căn(1/10001) < căn(1/10000) = 1/100 (đpcm)
đặt A= \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
B=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{10000}{10001}\)
Lấy A.B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}=\frac{1}{10001}\)
mặt khác
Ta có
\(\frac{1}{2}< \frac{2}{3}\\\)
\(\frac{3}{4}< \frac{4}{5}\)
....
\(\frac{9999}{10000}< \frac{10000}{10001}\)
=> A<B
=> A.A<A.B
=>A2<\(\frac{1}{10001}< \frac{1}{10000}\)
=>A<\(\sqrt{\frac{1}{10000}}=\frac{1}{100}\)
Vậy \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)<\(\frac{1}{100}\)
ĐPCM
Ta có :
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
Giá trị không đổi khi cả tử và mẫu cùng nhân với 2, ta được :
\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)
\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{19}\right)=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)
\(A=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}=\frac{1}{2}.\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{19}\right)=\frac{9}{19}\)
\(E=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{9999}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\right)\)
\(=50-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
\(=50-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=50-\frac{1}{2}.\left(1-\frac{1}{101}\right)=50-\frac{1}{2}.\frac{100}{101}=50-\frac{50}{101}=\frac{5000}{101}\)
Cảm ơn bạn nha !!!