Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
M=1/10 + 1/15 + 1/21 +....+ 1/120
M=2/20 +2/30+2/42+....+2/240
M=2/4.5 + 2/5.6 + 2/6.7 +.....+ 2/15.16
M=2.(1/4.5 +......+ 1/15.16)
M=2.(1/4 -1/5 +1/5 - 1/6 +.....+ 1/15 - 1/16)
M=2.(1/4 - 1/16)
M=2.(4/16 - 1/16)
M=2. 3/16
M=6/16=3/8
Có 1/3 = 8/24 < 9/24 = 3/8 =>1/3<M
Có 1/2 = 4/8>3/8 =>1/2 >M
=> 1/3 < M < 1/2
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{1}{2^2}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\left(2-\frac{1}{7}\right)=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\).
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)
=> x + 1 = 1993
=> x = 1993 - 1
=> x = 1992
\(\frac{7}{3}:\left(4.x-1\right)^2-\frac{1}{4}=\frac{1}{3}\)
\(\frac{7}{3}:\left(4.x-1\right)^2=\frac{1}{3}+\frac{1}{4}\)
\(\frac{7}{3}:\left(4.x-1\right)^2=\frac{7}{12}\)
\(\left(4.x-1\right)^2=\frac{7}{3}:\frac{7}{12}\)
\(\left(4.x-1\right)^2=4\)
\(\left(4.x-1\right)^2=2^2\)
\(4.x-1=2\)
\(4.x=2+1\)
\(4.x=3\)
\(x=3:4\)
\(x=0,75\)
H = 2012 - 1 - ( \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+99}\))
= 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{\left(99+1\right).\left[\left(99-1\right):1+1\right]:2}\)
= 2011 - ( \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\))
= 2011 - 2.( \(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\))
= 2011 - 2.(\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\))
= 2011 - 2.( \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\))
= 2011 - 2.(\(\frac{1}{2}-\frac{1}{100}\)) = 2011 - 2.\(\frac{49}{100}\)= 2011 - \(\frac{49}{50}\)= \(\frac{100501}{50}\)
\(H=2012-\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}\right)\)
\(=2012-\left(1+\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{99\left(99+1\right):2}\right)\)
\(=2012-\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\right)\)
\(=2012-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2}{99.100}\right)\)
\(=2012-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2012-2\left(1-\frac{1}{100}\right)\)
\(=2012-2\cdot\frac{99}{100}\)
\(=2012-\frac{99}{50}\)
\(=\frac{100501}{50}\)
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)