\(y=\frac{1}{1^4+1^2+1}+\frac{1}{2^4+2^2+1}+\frac{1}{3^4+3^2+1}+............+\frac{2014}{2014^4+2014^2+1}\)
Những câu hỏi liên quan
Rút gọn biểu thức : \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+2}+\frac{3}{3^4+3^2+3}+...+\frac{2014}{2014^4+2014^2+2014}\)
Rút gọn biểu thức: \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+...+\frac{2014}{2014^4+2014^2+1}\)
Rút gọn biểu thức: \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+.....+\frac{2014}{2014^4+2014^2+1}\)
Rut gon bieu thuc
\(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+...+\frac{2014}{2014^4+2014^2+1}\)
Áp dụng a/(a^4+a^2+1)=1/2.(1/(a^2-a+1)-1/(a^2+a+1)) ta được
A=1/2.(1/(1^2-1+1)-1/(1^2+1+1)+1/(2^2-2+1)-1/(2^2+2+10)+...+1/(2014^2-2014+1)-1/(2014^2+2014+1))
A=1/2.(1-1/(2014^2+2014+1))
A=-2029105/4058211
(CHẮC CHẮN ĐÚNG)
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Rút gọn biểu thức \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+...+\frac{2014}{2014^4+2014^2+1}\)
ở mẫu n4+n2+1=(n2+n+1)(n2-n+1)
\(\frac{2n}{n^4+n^2+1}=\frac{\left(n^2+n+1\right)-\left(n^2-2+1\right)}{\left(n^2-n+1\right)\left(n^2+n+1\right)}\)
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Tính:
A= 2014 + \(\frac{2014}{1+2}+\frac{2014}{1+2+3}+\frac{2014}{1+2+3+4}+........+\frac{2014}{1+2+3+4+.....+2013}\)
\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)
\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)
\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2014}\right)\)
\(A=2.2014.\frac{2013}{2014}\)
\(A=\frac{2.2014.2013}{2014}\)
\(A=2.2013\)
\(A=4026\)
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Tính tổng \(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+\frac{2014}{1+2+3+4}\)\(+...+\frac{2014}{1+2+3+...+10000}\)
\(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+...+\frac{2014}{1+2+3+...+10000}\)
\(S=\frac{2014}{\frac{1.2}{2}}+\frac{2014}{\frac{2.3}{2}}+\frac{2014}{\frac{3.4}{2}}+...+\frac{2014}{\frac{10000.10001}{2}}\)
\(S=\frac{4028}{1.2}+\frac{4028}{2.3}+\frac{4028}{3.4}+...+\frac{4028}{10000.10001}\)
\(S=4028\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10000.10001}\right)\)
\(S=4028\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10001-10000}{10000.10001}\right)\)
\(S=4028\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10000}-\frac{1}{10001}\right)\)
\(S=4028\left(1-\frac{1}{10001}\right)=\frac{40280000}{10001}\)
A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+.....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2014}+\frac{1}{2015}}\)=
Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
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toán lớp 8 dễ quá vậy
A=2015
hình như thế
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\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+\frac{2010}{5}+....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{2014}+\frac{1}{2015}}\)
Trình bày tự luận giúp mình nha !
Khẩn cấp đó
ở tử số ta làm thế này
\(TS=\left(1+\frac{1}{2014}\right)+\left(1+\frac{1}{2013}\right)+\left(1+\frac{1}{2012}\right)+...+\left(1+\frac{2013}{2}\right)\)
\(TS=2015\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+...+\frac{1}{2}\right)\)
\(\frac{TS}{MS}=2015\)
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