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Ta có: \(B=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
\(B=1+\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+\left(\frac{3}{2014}+1\right)+...+\left(\frac{2015}{2}+1\right)\)
\(B=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)
\(B=2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}\)
\(\Rightarrow\frac{A}{B}=\frac{1}{2017}.\)
Chúc bạn học tốt!
Này Vũ Minh Tuấn, mk cũng có 1 bài cũng gần giống như thế này nhưng khác 1 tí cậu giải giúp mk vs
\(N=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
\(N=1+\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+\left(\frac{3}{2014}+1\right)+...+\left(\frac{2015}{2}+1\right)\)
\(N=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)
\(N=2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}=\frac{1}{2017}\)
c) <=> \(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1\)\(+\frac{x+3}{2014}+1\)= \(\frac{x+4}{2013}+1+\frac{x+5}{2012}+1\)\(+\frac{x+6}{2011}\)
<=> \(\frac{x+1+2016}{2016}+\frac{x+2+2015}{2015}+\frac{x+3+2014}{2014}\) \(=\frac{x+4+2013}{2013}+\frac{x+5+2012}{2012}+\frac{x+6+2011}{2011}\)
<=> \(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
<=> \(\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
vì \(\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)khác 0
=> \(x+2017=0\) => \(x=-2017\)
Vậy \(S=\left\{-2017\right\}\)
vì 2>0\(\Rightarrow|x-2014|+|x-2015|+|x-2016|>0\)
\(\Rightarrow|x-2014|+|x-2015|+|x-2016|\)
\(\Rightarrow x-2014+x-2015+x-2016=2\)
\(\Rightarrow x+x+x-2014-2015-2016=2\Rightarrow3x-6045=2\)
\(\Rightarrow3x=6047\Rightarrow x=6047:3=\frac{6047}{3}\)
bạn jj vừa trả lời ơi, cho mik hỏi tí là vì sao bn suy ra đc dòng 3
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\left(\frac{x-1}{2016}+1\right)+\left(\frac{x-2}{2015}+1\right)=\left(\frac{x-3}{2014}+1\right)+\left(\frac{x-4}{2013}+1\right)\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2014}\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x+2017}{2014}-\frac{x+2017}{2013}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(\Leftrightarrow x-2017=0\)
\(\Leftrightarrow x=2017\)
\(\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016} \)
\(\Leftrightarrow\frac{x+4}{2013}+1+\frac{x+3}{2014}+1-\frac{x+2}{2015}-1-\frac{x+1}{2016}-1=0\)
\(\Leftrightarrow\frac{x+2014+2013}{2013}+\frac{x+3+2014}{2014}-\frac{x+2+2015}{2015}-\frac{x+1+2016}{2016}=0\)
\(\Leftrightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2017=0\) ( vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\)>0)
\(\Leftrightarrow x=2017\)
A = 22016 - 22015 - 22014 - ... - 2 - 20
A = 22015 - 22014 - ... - 2 - 20
...
A = 2 - 20
A = 1