Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) Ta có: \(S=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\)
\(=\frac{1}{2}-\frac{1}{300}=\frac{149}{300}< \frac{200}{300}=\frac{2}{3}\)
hay \(S< \frac{2}{3}\)(1)
Ta có: \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
nên \(\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(vì mỗi ngoặc trên đều có 100 phân số có tử là 1)
\(\Leftrightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}>\frac{1}{200}\cdot100+\frac{1}{300}\cdot100\)
\(\Leftrightarrow Q>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
mà \(\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)
nên \(Q>\frac{2}{3}\)
hay \(\frac{2}{3}< Q\)(2)
Từ (1) và (2) suy ra S<Q
(1/12+3 1/6-30,75).x -8 = (3/5+0,415+1/200):0,01
(1/12+19/6-123/4).x-8=(3/5+83/200+1/200):1/100
-55/2.x-8=51/50:1/100
-55/2.x-8=102
-55/2.x=102+8=110
x=110:-55/2=-4
a) Ta có:
\(A=-3\cdot7\cdot\left(-2\right)\cdot\left(-13\right)\)
\(A=-21\cdot26\)
\(A=-546\)
\(B=-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot5\)
\(B=2\cdot12\cdot5\)
\(B=2\cdot60\)
\(B=120\)
Mà: \(120>-546\)
\(\Rightarrow B>A\)
1/ So sánh A với \(\frac{1}{4}\)
Có \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.........+\frac{1}{2014.2015.2016}\)
\(A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-.......+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(A=\frac{1}{1.2}-\frac{1}{2015.2016}=\frac{1}{2}-\frac{1}{2015.2016}\)
Vậy \(A>\frac{1}{4}\)
a) ta có công thức \(\frac{a}{n.\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)
ta có \(N=\frac{5^2}{5.10}+\frac{5^2}{10.15}+...+\frac{5^2}{2005.2010}\)
\(N=5\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2005.2010}\right)\)
\(N=5\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)(sử dụng quy tắc dấu ngoặc)
\(N=5\left[\frac{1}{5}-\left(\frac{1}{10}-\frac{1}{10}\right)-\left(\frac{1}{15}-\frac{1}{15}\right)-...-\left(\frac{1}{2005}-\frac{1}{2005}\right)-\frac{1}{2010}\right]\)
\(N=5\left[\frac{1}{5}-0-0-...-0-\frac{1}{2010}\right]\)
\(N=5\left[\frac{1}{5}-\frac{1}{2010}\right]\)
\(N=5.\frac{401}{2010}\)
\(N=\frac{401}{402}\)
b) \(M=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)
ta thấy \(\frac{1}{11}=\frac{1}{11}\)
\(\frac{1}{12}<\frac{1}{11}\)
\(\frac{1}{13}<\frac{1}{11}\)
.................
\(\frac{1}{20}<\frac{1}{11}\)
\(\Rightarrow M=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}<\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}\)(có 10 phân số \(\frac{1}{11}\))
\(\Rightarrow\frac{1+1+1...+1}{11}\)
\(=\frac{10}{11}\)
ta có \(\frac{10}{11}=\frac{4020}{4422}\)(1)
\(\frac{401}{402}=\frac{4411}{4422}\)(2)
từ (1)và (2)\(\Rightarrow\frac{4020}{4422}<\frac{4411}{4422}\Leftrightarrow\frac{10}{11}<\frac{401}{402}\)
Vì \(M<\frac{10}{11}<\frac{401}{402}=N\left(3\right)\)
Từ \(\left(3\right)\Leftrightarrow M