Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+.....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)
a)Đặt A=Tổng trên, ta có:
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2A=2+1+...+\frac{1}{2^{99}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(A=2-\frac{1}{2^{100}}\)
b)có đứa làm rồi
c)Đặt C=Tổng trên
\(3C=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(3C=1+\frac{1}{3}+...+\frac{1}{3^{299}}\)
\(3C-C=\left(1+\frac{1}{3}+...+\frac{1}{3^{299}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(2C=1-\frac{1}{3^{300}}\)
\(C=\frac{1-\frac{1}{3^{300}}}{2}\)
A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
B = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
Bạn tham khảo ở link này nhé :
Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến
a,1/102+1/112+1/122+...+1/1002<1/9.10+1/10.11+1/11.12+...+1/99.100=1/9-1/10+1/10-1/11+...+1/99-1/100
=1/9-1/100=91/900<3/4
Vậy 1/102+1/112+1/122+...+1/1002<3/4
b,1/22+1/32+1/42+...+1/1002<1/1.2+1/2.3+1/3.4+...+1/99.100=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100=99/100
Vậy 1/22+1/32+1/42+...+1/1002<99/100
c,1/22+1/32+1/42+...+1/1002<1/22+(1/2.3+1/3.3+...+1/99.100)=1/4+(1/2-1/3+1/3-1/4+...+1/99-1/100)
=1/4+(1/2-1/100)=1/4+49/100=74/100<3/4=75/100
Vậy 1/22+1/32+1/42+...+1/1002<3/4
M=1/2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19
M=1/2-1/2+1/7-1/7+1/5-1/5+1/13-1/13+1/8-1/8+1/19
M=1/2-1/19
M=17/38
\(G=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..............+\frac{1}{3^{100}}\)
\(3G=1+\frac{1}{3}+\frac{1}{3^2}+...............+\frac{1}{3^{99}}\)
\(3G-G=\left(1+\frac{1}{3}+\frac{1}{3^2}+..........+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...............+\frac{1}{3^{100}}\right)\)
\(2G=1-\frac{1}{3^{100}}\)
\(\Rightarrow G=\left(1-\frac{1}{3^{100}}\right):2\)