rut gon bieu thuc
1/(1^4+1^2+1)+2/(2^4+2^2+1)+3/(3^4+3^2+1)+...+2014/(2014^4+2014^2+2014)=...
{giup minh vs}
Những câu hỏi liên quan
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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tinh nhanh bieu thuc sau
a) x^6-20x^5-20x^4-20x^4-20x^3-20x^2-20x+1992 voi x= 21
b)x^7-26x^6+27x^5-47x^4-77x^3+50x^2+x+1989 tai x = 25
c)4029-2015x+2015x^2-....+2015x^2014-x^2015 tai x= 2014
d)A= (x-1/x-2015+x-2/x-2014+x-3/x-2013+....+x-2014/x-2+x-2015/x-1):(1/2 +1/3+1/4+....+1/2016) tai x=2016 dau / o day la phan do nha!!
xin nho cac ban giai giup cho minh! cam on nhieu!!!!!!!! minh dang can ket qua gap
Chưng minh rằng :
C= 1-\(\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2014^2}>\frac{1}{2014}\)
Giup mk vs nha!<3<3<3
A=2014+[2014:(1+2)]+[2014:(1+2+3)]+[2014:(1+2+3+4)]+...++[2014:(1+2+3+...+2013)]
Rút gọn biểu thức : \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+2}+\frac{3}{3^4+3^2+3}+...+\frac{2014}{2014^4+2014^2+2014}\)
Chứng minh: (1/4+2/4^2+3/4^34/4^4+........+2014/4^2014)<1/2
Rut gon bieu thuc B=3/2-(3/2)^2+(3/2)^3-.........(3/2)^2014
Ta có :\(B=\frac{3}{2}-\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3-....-\left(\frac{3}{2}\right)^{2014}\)
\(\frac{3}{2}B=\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4-...+\left(\frac{3}{2}\right)^{2014}-\left(\frac{3}{2}\right)^{2015}\)
\(\frac{3}{2}B+B=\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^3+..+\left(\frac{3}{2}\right)^{2014}-\left(\frac{3}{2}\right)^{2015}\) \(+\frac{3}{2}-\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3-...-\left(\frac{3}{2}\right)^{2014}\)
\(\frac{5}{2}B=\frac{3}{2}-\left(\frac{3}{2}\right)^{2015}\)
\(B=\frac{\frac{3}{2}-\left(\frac{3}{2}\right)^{2015}}{\frac{5}{2}}\)
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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1) 1/2 + 1/3 + 1/4 + ... + 1/2013 + 1/2014
2) 2014 + 2013/2 + 2012/3 + 2011/4 + ... + 2/2013 + 1/2014
