Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt N=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+......+\(\frac{2x}{x\left(x+1\right)}\)
N=\(\frac{2}{6}\)+\(\frac{2}{12}\)+\(\frac{2}{20}\)+.....+\(\frac{2x}{x\left(x+1\right)}\)
N=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)+\(\frac{2}{4.5}\)+.....+\(\frac{2x}{x\left(x+1\right)}\)
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
a) \(y+2\frac{3}{4}=5\frac{2}{3}\)
\(y+\frac{11}{4}=\frac{17}{3}\)
y = 35/12
b) \(y-1\frac{4}{5}=3\frac{2}{7}\)
y - 9/5 = 23/7
y = 178/35
\(a,y+2\frac{3}{4}=5\frac{2}{3}\)
\(\Rightarrow y+\frac{11}{4}=\frac{17}{3}\)
\(\Rightarrow y=\frac{17}{3}-\frac{11}{4}\)
\(\Rightarrow y=\frac{35}{12}\)
\(b,y-1\frac{4}{5}=3\frac{2}{7}\)
\(\Rightarrow y=3\frac{2}{7}-1\frac{4}{5}\)
\(\Rightarrow y=\frac{52}{35}\)
\(=\frac{-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+1-...-\frac{92}{100}+1}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{8\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}}\)
= 8
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)
=> x + 1 = 1993
=> x = 1993 - 1
=> x = 1992
\(-\frac{3}{2}+-\frac{3}{10}\)
\(=-\frac{3.5}{2.5}+-\frac{3}{10}\)
\(=-\frac{15}{10}+-\frac{3}{10}\)
\(=\frac{\left(-15\right)+\left(-3\right)}{10}\)
\(=-\frac{18}{10}=-\frac{9}{5}\)
= -9/5 nha cj