Tính\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\)
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12 tháng 6 2018
2.
a) Ta có:
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b) Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)
Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)
Vậy, x = -2004
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
28 tháng 2 2020
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100}\)
\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}...+\frac{1}{99}\)
\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow2S-S=S=2-\frac{1}{2^{100}}=\frac{2^{101}}{2^{100}}-\frac{1}{2^{100}}=\frac{2^{101}-1}{2^{100}}\)
T
16 tháng 3 2018
Đăng từ bài thôi bạn à!
a) Áp dụng công thức: \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3}-\frac{1}{4}\)
..............................
\(\frac{1}{n^2}< \frac{1}{n-1}-\frac{1}{n}\)
___________________________________________
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}=\frac{1}{n+1}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\) (đpcm)
NN
21 tháng 9 2020
Đặt: \(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+........+2019}\)
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .............. ; \(1+2+3+......+2019=\frac{2019.2020}{2}\)
\(\Rightarrow B=\frac{2}{2}+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+........+\frac{1}{\frac{2019.2020}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{2019.2020}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2019.2020}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2.\left(1-\frac{1}{2020}\right)=2.\frac{2019}{2020}=\frac{2019}{1010}\)
\(\Rightarrow A=\frac{2.2019}{\frac{2019}{1010}}=2.1010=2020\)
AK
10 tháng 4 2018
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)
Chúc bạn học tốt !!!
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 \(x=2015\)
5 tháng 8 2017
\(\frac{\frac{2}{3}+\frac{1}{4}-\frac{3}{5}}{\frac{2}{3}-\frac{1}{4}+\frac{3}{5}}=\frac{\frac{2}{3}+\frac{1}{4}-\frac{3}{5}}{\frac{2}{3}-\left(\frac{1}{4}-\frac{3}{5}\right)}=\frac{\frac{2}{3}-\frac{7}{20}}{\frac{2}{3}+\frac{7}{20}}=\frac{\frac{19}{60}}{\frac{61}{60}}=\frac{19}{60}\times\frac{60}{61}=\frac{19}{61}\)
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~
5 tháng 8 2017
ta thấy có thể giản ước 2/3 = 2/3 ,1/4=1/4,3/5=3/5
=> phép tính trên bằng 1
8 tháng 8 2019
a) \(\frac{-1}{2}+\frac{-1}{9}-\frac{-3}{5}+\frac{1}{2006}-\frac{-2}{7}-\frac{7}{18}+\frac{4}{35}\)
\(=\left(\frac{-1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-9-2-7}{18}\right)+\left(\frac{21+4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-18}{18}\right)+\left(\frac{25}{35}\right)+\frac{1}{2006}\)
\(=\left(-1\right)+\frac{5}{7}+\frac{1}{2006}\)\(=\frac{-4005}{14042}\)
b) \(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{2007}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)
\(=\left(\frac{1}{3}+\frac{1}{2007}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)\)
\(=\left(\frac{669}{2007}+\frac{1}{2007}-\frac{446}{2007}\right)-\left(\frac{27}{36}+\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)\)
\(=\frac{224}{2007}-\frac{28}{36}+\frac{10}{15}\)
\(=\frac{224}{2007}-\frac{1561}{2007}+\frac{1338}{2007}\)\(=\frac{1}{2007}\)
mình đang nói cá muối nha bạn
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}.\)
\(\frac{1}{A}=2+3+4+....+100\)
Dãy trên có tất cả số số hạng là :
( 100 - 2 ) : 1 + 1 = 99 ( số hạng )
Tổng của dãy trên là :
( 100 + 2 ) x 99 : 2 = 5049
Thay vào ta có :
\(\frac{1}{A}=5049\)
\(\Rightarrow A=\frac{1}{5049}\)
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