Câu 1 Tính:
a. M= \(\frac{363636.37-373737.36}{1+2+3+...+2013}\)
b, B= (\(\frac{1}{11}+\frac{1}{21}+\frac{1}{31}+\frac{1}{41}+\frac{1}{51}\)) x (\(\frac{123}{738}-\frac{1342}{2013}-\frac{-1007}{2014}\))
c, D= \(\frac{12}{6.7}+\frac{12}{7.22}+\frac{12}{22.15}+\frac{12}{15.38}+...+\frac{12}{97.202}\)
K
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Những câu hỏi liên quan
23 tháng 7 2017
a , Ta có : \(1-\frac{54}{59}=\frac{5}{59}\) \(=\frac{50}{590}\) ; \(1-\frac{541}{591}=\frac{50}{591}\)
Vì \(\frac{50}{590}>\frac{50}{591}\)nên \(\frac{54}{59}< \frac{541}{591}\)
S
6 tháng 7 2018
a, ĐK: \(x+1\ge0\Leftrightarrow x\ge-1\)
Ta có: |3-2x|=x+1
=>\(\orbr{\begin{cases}3-2x=x+1\\3-2x=-x-1\end{cases}\Rightarrow\orbr{\begin{cases}x+2x=3-1\\-x+2x=3+1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=2\\x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\left(tmđk\right)\\x=4\left(tmđk\right)\end{cases}}}\)
Vậy x=2/3 hoặc x=4
b, Xét VP ta có: \(\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}+1\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(x=\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}=2014\)
Vậy x=2014
S
15 tháng 7 2017
Ta có: \(M=\frac{2014^2+1^2}{2014.1}+\frac{2013^2+2^2}{2013.2}+\frac{2012^2+3^2}{2012.3}+...+\frac{1008^2+1007^2}{1008.1007}\)
\(=\frac{2014}{1}+\frac{1}{2014}+\frac{2013}{2}+\frac{2}{2013}+\frac{2012}{3}+\frac{3}{2013}+...+\frac{1008}{1007}+\frac{1007}{1008}\)
\(=\frac{2014}{1}+\frac{2013}{2}+...+\frac{1}{2014}\)
\(=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)\)
\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}}=2015\)
21 tháng 4 2015
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
HT
18 tháng 9 2015
\(S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}}\)
Xét mẫu:
\(\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}\)
= \(\left(1+\frac{2013}{2}\right)+\left(1+\frac{2012}{3}\right)+...+\left(1+\frac{1}{2014}\right)+1\)
= \(\frac{2014}{2}+\frac{2014}{3}+....+\frac{2014}{2013}+\frac{2014}{2014}\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)
\(\Rightarrow S=\frac{1}{2014}\)
\(D=12\cdot\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{22}+...+\frac{1}{97}-\frac{1}{202}\right)\)
\(D=12\cdot\left(\frac{1}{6}-\frac{1}{202}\right)\)
\(D=12\cdot\frac{49}{303}\)
\(D=\frac{588}{303}\)