Tìm x biết
[x+2]+[x+3]+...+[x+2013]=2013+2014
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\(x-2014-\frac{2015}{2013}+x-2013-\frac{2015}{2014}+x-2014-\frac{2013}{2015}=3\)
\(\Rightarrow\left(x+x+x\right)+\left(-2014-2014\right)-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)
\(3x-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)
\(3x=3+2013+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2013}{2015}\)
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<=> \(\frac{\left(x+2014\right)}{2011}+1+\frac{\left(x+2013\right)}{2012}+1=\frac{\left(x+2012\right)}{2013}+1+\frac{\left(x+2011\right)}{2014}+1\)
\(\Rightarrow\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}=\frac{\left(x+4025\right)}{2013}+\frac{\left(x+4025\right)}{2014}\)
=> \(\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}-\frac{\left(x+4025\right)}{2013}-\frac{\left(x+4025\right)}{2014}=0\)
=> \(\left(x+4025\right)\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack=0\)
vì \(\left(\frac{1}{2011}+\frac{1}{2012}\right)>\left(\frac{1}{2013}+\frac{1}{2014}\right)\)
=> \(\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack>0\) hay ≠0
=> \(x+4025=0\)
\(x=-4025\)
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
=> \(\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)
=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)
=> \(\left(x+2016\right).\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Vì \(\frac{1}{2012}>\frac{1}{2014};\frac{1}{2013}>\frac{1}{2015}\)
=> \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
=> \(x+2016=0\)
=> \(x=-2016\)
\(\left[x+2\right]+\left[x+3\right]+\left[x+4\right]+.....+\left[x+2013\right]=2013+2014\)
\(\left[x+x+....+x\right]+\left[2+3+4+......+2013\right]=4027\)
\(2012x+2027090=4027\)
\(2012x=4027-2027090\)
\(2012x=-2023063\)
\(x=-2023063:2012\)
\(x=-\frac{2023063}{2012}\)
[x+2]+[x+3]+....+[x+2013] = 2013+2014
=> x+2+x+3+...+x+2013 = 4027
=> (x+x+x+..+x) +(2+3+4+...+2013) = 4027
=> 2012x + 2027090 = 4027
=> 2012x = -2023063
=> x = \(\frac{-2023063}{2012}\)