Tìm x thuộc N, biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2012}\)
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Những câu hỏi liên quan
8 tháng 1 2017
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
ML
29 tháng 6 2015
1/
\(1+\frac{2014}{2}+...+\frac{4024}{2012}=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{2012}{2012}\right)\)
\(=2012+2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
Phương trình đã cho tương đương:
\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
\(\Leftrightarrow503x=2012\)
\(\Leftrightarrow x=4\)
2/
\(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}+\left(1+\frac{1}{57}\right)-2-\frac{7}{3}+\frac{8}{4}=5x\)
\(\Leftrightarrow\)\(5x=\frac{17}{3}\Leftrightarrow x=\frac{17}{15}\)
3/
Ta có: \(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{n\left(n+2\right)}\right)\)\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}.......\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(=2.\frac{n+1}{n+2}
NH
0
2 tháng 1 2017
Anh chỉ giải câu a thôi, câu b anh thấy nó bình thường mà.
Cộng vào mỗi phân số thêm 1 đơn vị được:
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\).
Tới đây tự làm tiếp nhá.
27 tháng 1 2020
Xét \( A = 1 + \dfrac{{2014}}{2} + \dfrac{{2015}}{3} + ... + \dfrac{{4023}}{{2011}} + \dfrac{{4024}}{{2012}}\\ \)
\(\Rightarrow A - 2012 = \left( {\dfrac{{2014}}{2} - 1} \right) + \left( {\dfrac{{2015}}{3} - 1} \right) + ... + \left( {\dfrac{{4024}}{{2012}} - 1} \right)\\ \Rightarrow A - 2012 = \dfrac{{2012}}{2} + \dfrac{{2012}}{3} + ... + \dfrac{{2012}}{{2012}}\\ \Rightarrow A - 2012 = 2012\left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow A = 2012\left( {1 + \dfrac{1}{2} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)503x = 2012\left( {1 + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow x = \dfrac{{2012}}{{503}} = 4 \)
8 tháng 1 2017
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow2^x\cdot1+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=480\)
\(\Rightarrow2^x\left(1+2^1+2^2+2^3\right)=480\)
\(\Rightarrow2^x\cdot15=480\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
8 tháng 1 2017
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(\frac{2011}{2}+1\right)+...+\left(\frac{2}{2011}+1\right)+\left(\frac{1}{2012}+1\right)+1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow x=2013.\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}\)
\(\Rightarrow x=2013\)
Vậy \(x=2013\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)
\(\Rightarrow A=\left(\frac{1}{2}-\frac{1}{x+1}\right):\frac{1}{2}\)
Theo bài ra ta có:
\(\left(\frac{1}{2}-\frac{1}{x+1}\right):\frac{1}{2}=\frac{2011}{2013}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2013}{4026}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
=> x + 1 = 2013
=> x = 2013 - 1
=> x = 2012 \(\in\) N
Vậy x = 2012
Đặt S=1/3+1/6+1/10+..........+2/x(x+1)
1/2S=1/2[1/3+1/6+1/10+...+2/x(x+1)]
1/2S=1/6+1/12+1/20+......1/x(x+1)
1/2S=1/2.3+1/3.4+1/4.5+.....+1x(x+1)
1/2S=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1
1/2S=1/2-1/x+1
Vì S=2011/2013
suy ra (1/2-1/x+1):1/2=2011/2013
(1/2-1/x+1).2=2011/2013
1/2-1/x+1=2011/2013:2
1/2-1/x+1=2011/4026
1/x+1=1/2-2011/4026
1/x+1=1/2013
suy ra x+1=2013
x=2013-1
x=2012