Tính S=\(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+......+\frac{2011}{2011^4+2011^2+1}\)
K
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G
4 tháng 3 2018
Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)
\(L=1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(L=\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+..+\frac{2012}{2011}\)
\(L=2012\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2012}\right)\)
Hay: \(P=\frac{1}{2012}\)
CV
6 tháng 10 2016
Câu 1) Ta có\(a^3+2b^2-4b+3=0\Leftrightarrow a^3=-2.\left(b-1\right)^2-1\)\(\le-1\Rightarrow a^3\le-1\Rightarrow a\le-1\Rightarrow a^2\ge1\)
\(\Rightarrow\hept{\begin{cases}a^2\ge1\\a^2b^2\ge b^2\end{cases}}\)\(\Rightarrow a^2+a^2b^2-2b\ge1+b^2-2b\)\(\Leftrightarrow\left(b-1\right)^2\le0\)
Mà \(\left(b-1\right)^2\ge0\)với mọi b nên \(\left(b-1\right)^2=0\)\(\Rightarrow b=1\)
Thay b=1 vào 2 pt ban đầu được \(\hept{\begin{cases}a^3+2-4+3=0\\a^2+a^2-2=0\end{cases}}\)<=> a=1(tm)
Vậy (a,b)=(1;1)
Câu 2 bạn xem ở đây nhé http://olm.vn/hoi-dap/question/716469.html
9 tháng 8 2017
Ta có :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....+\left(\frac{1}{2011}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+....+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}\right)}\)
\(\frac{1}{2012}\)
55
19 tháng 6 2019
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
19 tháng 6 2019
\(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{1}{2011}\)
\(=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....\left(\frac{1}{2011}+1\right)+1\)
\(=\frac{2012}{2}+\frac{2012}{3}+\frac{2012}{4}+.....+\frac{2012}{2011}+\frac{2012}{2012}\)
\(=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}\right)\)
Thay vào,rút gọn là ra
10 tháng 11 2015
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}}\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
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\(=\frac{2023066}{4046133}\)