So sánh A với 1 biết rằng: A =\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{35.37}\)
Mik có chuyện gấp phiền các bạn mong các bạn giúp mình
K
Khách
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Những câu hỏi liên quan
2 tháng 5 2016
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{21.23}\)
\(=5-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+\frac{5}{5}-\frac{5}{7}+...+\frac{5}{21}-\frac{5}{23}\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=5\left(1-\frac{1}{23}\right)\)
\(=5.\frac{22}{23}\)
\(=\frac{110}{23}\)
2 tháng 5 2016
\(A=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{21}-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\frac{22}{23}\)
\(A=\frac{55}{23}\)
NT
7 tháng 5 2018
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
\(=1-\left(\frac{2}{1.3}-\frac{2}{3.5}-...-\frac{2}{2005-2007}\right)\)
\(=1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(=1-\left[1+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{5}+\frac{1}{5}\right)+...+\left(-\frac{1}{2005}+\frac{1}{2005}\right)-\frac{1}{2007}\right]\)
\(=1-\left(1+0+0+...+0-\frac{1}{2007}\right)\)
\(=1-\left(1-\frac{1}{2007}\right)\)
\(=1-1+\frac{1}{2007}\)
\(=0+\frac{1}{2007}\)
\(=\frac{1}{2007}\)
Ai thấy tớ đúng k nha
S
15 tháng 5 2017
Đặt A = \(1-\frac{2}{1.3}-\frac{2}{3.5}-.....-\frac{2}{2005.2007}\)
= \(1-\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2005.2007}\right)\)
=\(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2005}-\frac{1}{2007}\right)\)
= \(1-\left(1-\frac{1}{2017}\right)\)
=\(1-1+\frac{1}{2017}\)
=\(0+\frac{1}{2017}\)
=\(\frac{1}{2017}\)
5 tháng 5 2018
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)
LA
6 tháng 3 2018
\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)
\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)
\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)
20 tháng 5 2020
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XO
8 tháng 7 2021
\(\frac{1}{1.3}+\left|\frac{-1}{3.5}\right|+\left|\frac{-1}{5.7}\right|+...+\left|\frac{-1}{49.51}\right|=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\left(1-\frac{1}{51}\right)=\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)
HL
8 tháng 7 2021
\(\frac{1}{1.3}+\left|-\frac{1}{3.5}\right|+\left|-\frac{1}{5.7}\right|+...+\left|-\frac{1}{49.51}\right|\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49\cdot51}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{50}{51}\)
\(=\frac{25}{51}\)
10 tháng 5 2018
\(\text{Câu 1 :}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{1}-\frac{1}{13}\)
\(=\frac{12}{13}\)
\(\text{Câu 2 :}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{35.37}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{1}-\frac{1}{37}<1\text{ Vậy }A<1\)
Monkey D.Luffy điêu vừa thui mới học lớp 6 mà ông cũng nói dễ