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Xét N có:
\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta các số hạng của M và N có:
\(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\) (1)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\) (2)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\) (3)
Từ (1);(2);(3) => M > N
5\(\frac{8}{17}\) : x - (\(\frac{8}{17}\)) : x + 3\(\frac{1}{17}\): 17\(\frac13\) = \(\frac{4}{17}\)
(5\(\frac{8}{17}\) - \(\frac{8}{17}\)) : x + \(\frac{52}{17}\) : \(\frac{52}{3}\) = \(\frac{4}{17}\)
5:x + \(\frac{52}{17}\times\) \(\frac{3}{52}\) = \(\frac{4}{17}\)
5 : x + \(\frac{3}{17}\) = \(\frac{4}{17}\)
5 : x = \(\frac{4}{17}\) - \(\frac{3}{17}\)
5 : x = \(\frac{1}{17}\)
x = 5 : \(\frac{1}{17}\)
x = 5 x 17
x = 85
Vậy x = 85
1/1.4 + 1/4.7 + ...+1/x(x+3) = 6/19
3/1.4 + 3/4.7 +..+3/x(x+3) = 18/19
1/1 - 1/4+ 1/4 - 1/7 +...+1/x-1/(x+3) = 1 - 1/19
1- 1/(x+3) = 1 - 1/19
1/(x+3) = 1/19
x + 3 = 19
x = 19 - 3
x = 16
Vậy x = 16
Ta có
\(\frac{2014}{1}+\frac{2015}{2}+...+\frac{4026}{2013}=1+1+...+1+\left[\left(\frac{2014}{1}-1\right)+\left(\frac{2015}{2}-1\right)+...+\left(\frac{4026}{2013}-1\right)\right]\)
\(=2013+\left(\frac{2013}{1}+\frac{2013}{2}+...+\frac{2013}{2013}\right)=2013+2013\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)\) (1)
Ta kết hợp (1) và đề
=>\(\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)x+2013=2013+2013\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)\)
=> x=2013
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=\left(\frac{2014}{1}-1\right)+\left(\frac{2015}{2}-1\right)+...+\left(\frac{4025}{2012}-1\right)+\left(\frac{4026}{2013}-1\right)\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=\frac{2013}{1}+\frac{2013}{2}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=2013\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
\(\Rightarrow x=\frac{2013\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}=2013\)
Vậy x = 2013 thoả mãn đề bài.