1/2+1/3+1/4+1/5+..........1/2011
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20 tháng 8 2017
x2+y2+z2= xy+yz+zx.
=> 2x2+2y2+2z2-2xy-2yz-2zx=0
=> ( x-y)2+(y-z.)2+(z-x)2 =0
=> x=y=z=0
Thay x=y=z vào x2011+y2011+z2011=32012 ta được:
3.x2011=3.32011
=> x2011=32011
=> x=3 hoặc x = -3
Hay x=y=z=3 hoặc x=y=z=-3
20 tháng 8 2017
1) có bn giải rồi ko giải nữa
2) \(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2011^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2012^4+\frac{1}{4}\right)}\)
Với mọi n thuộc N ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2=\left(n^2-n+\frac{1}{2}\right)\left(n^2+n+\frac{1}{2}\right)\)
\(=\left[n\left(n-1\right)+\frac{1}{2}\right]\left[n\left(n+1\right)+\frac{1}{2}\right]\)
Áp dụng ta được :
\(A=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(2011.2012+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right).......\left(2012.2013+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{2012.2013+\frac{1}{2}}=\frac{1}{8100313}\)
NT
1-1/2+1/3-1/4+1/5-1/6+...+1/2011-1/2012 / 1006-1006/1007-1007/1008-1008/1009-...-2010/2011-2011/2012
0
28 tháng 5 2016
Xét biểu thức phụ : \(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
21 tháng 5 2018
Bài 3:
Ta có:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(...\)+\(\frac{1}{2010^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2009.2010}\)
Xét:\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{2009+2010}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)=\(1-\frac{1}{2010}\)<1
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2010^2}< 1\)
\(\)Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\)
CV
5 tháng 10 2016
Ta có: \(2.S=2.\left(\frac{1}{1^4+1^2+1}+...+\frac{2011}{2011^4+2011^2+1}\right)\)
Xét hạng tử tống quát: \(\frac{2.n}{n^4+n^2+1}=\frac{2.n}{\left(n^4+2n^2+1\right)-n^2}=\frac{\left(n^2+n+1\right)-\left(n^2-n+1\right)}{\left(n^2-n+1\right)\left(n^2+n+1\right)}\)\(=\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\)
Từ đó: \(\frac{2.1}{1^4+1^2+1}=\frac{1}{1}-\frac{1}{3}\)
\(\frac{2.2}{2^4+2^2+1}=\frac{1}{3}-\frac{1}{7}\)
.....
\(\frac{2.2011}{2011^4+2011^2+1}=\frac{1}{4042111}-\frac{1}{4046133}\)
Từ đó => 2.S= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+...+\frac{1}{4042111}-\frac{1}{4046133}\)=\(1-\frac{1}{4046133}\)=\(\frac{4046132}{4046133}\)
=> S\(=\frac{2023066}{4046133}\)